Conditional probability problem












4












$begingroup$


An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?



What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
then we need to find a conditional probability
$P(Xgeq 2 | Y)=P(X= 2,3,text{ or }4 | Y)=frac{P({Xgeq 2} cap Y)}{P(Y)}$
but then I don't know how to continue?



Any help...










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?



    What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
    then we need to find a conditional probability
    $P(Xgeq 2 | Y)=P(X= 2,3,text{ or }4 | Y)=frac{P({Xgeq 2} cap Y)}{P(Y)}$
    but then I don't know how to continue?



    Any help...










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?



      What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
      then we need to find a conditional probability
      $P(Xgeq 2 | Y)=P(X= 2,3,text{ or }4 | Y)=frac{P({Xgeq 2} cap Y)}{P(Y)}$
      but then I don't know how to continue?



      Any help...










      share|cite|improve this question











      $endgroup$




      An Internet search engine looks for a keyword in 9 databases, searching them in random order. Only 5 of these databases contain the given keyword. What is the probability that it will be found in at least 2 of the first 4 searched databases?



      What I tried is: Let X: the event that the keyword is to be found in at least 2 databases, and Y: the event that we search the first 4 databases,
      then we need to find a conditional probability
      $P(Xgeq 2 | Y)=P(X= 2,3,text{ or }4 | Y)=frac{P({Xgeq 2} cap Y)}{P(Y)}$
      but then I don't know how to continue?



      Any help...







      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 11 '11 at 0:03









      Michael Hardy

      1




      1










      asked Sep 10 '11 at 16:38









      KellyKelly

      363




      363






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.



          If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.



          So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.



          Add up the last three (and change to lowest terms) and you have your solution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
            $endgroup$
            – Henry
            Sep 10 '11 at 21:23












          • $begingroup$
            Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:48












          • $begingroup$
            So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:53










          • $begingroup$
            In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
            $endgroup$
            – Henry
            Sep 10 '11 at 22:09












          • $begingroup$
            Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 22:09



















          3












          $begingroup$

          Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.



          $$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          But



          $$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
          $$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          Thus the two answers give the same result.




          (Original answer.)



          I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.



          $$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.



          I'll let you calculate $P(X = 3)$ and $P(X = 4)$.



          (FYI, $X$ has what's called a hypergeometric distribution.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
            $endgroup$
            – Kelly
            Sep 10 '11 at 16:56










          • $begingroup$
            @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 17:05










          • $begingroup$
            I feel that there is something wrong above! Are we choosing 4 items and not 5?!
            $endgroup$
            – Kelly
            Sep 10 '11 at 17:06










          • $begingroup$
            @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 20:31



















          0












          $begingroup$

          Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.



          Let $c=4$ databases are chosen and $s$ databases have the keyword.



          The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.



          There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.



          Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.



          Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.



          Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.



          Hence, the required probability is:
          $$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
          frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
          frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$

          Note that you can also use the complement, i.e.:
          $$P(sge 2)=1-P(s=0)-P(s=1).$$



          Thus, in general, the formula is:
          $$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$



          Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Useful, after the answers from 7 years ago?
            $endgroup$
            – Did
            Jan 1 at 12:12










          • $begingroup$
            @Did, I tried to summarize and generalize the other answers.
            $endgroup$
            – farruhota
            Jan 1 at 12:21










          • $begingroup$
            How did you fall on this question and why did you decide to "reawaken" it?
            $endgroup$
            – Did
            Jan 1 at 12:24










          • $begingroup$
            @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
            $endgroup$
            – farruhota
            Jan 1 at 12:29










          • $begingroup$
            Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
            $endgroup$
            – o0omycomputero0o
            Jan 1 at 15:03











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f63356%2fconditional-probability-problem%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.



          If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.



          So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.



          Add up the last three (and change to lowest terms) and you have your solution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
            $endgroup$
            – Henry
            Sep 10 '11 at 21:23












          • $begingroup$
            Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:48












          • $begingroup$
            So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:53










          • $begingroup$
            In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
            $endgroup$
            – Henry
            Sep 10 '11 at 22:09












          • $begingroup$
            Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 22:09
















          3












          $begingroup$

          You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.



          If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.



          So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.



          Add up the last three (and change to lowest terms) and you have your solution.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
            $endgroup$
            – Henry
            Sep 10 '11 at 21:23












          • $begingroup$
            Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:48












          • $begingroup$
            So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:53










          • $begingroup$
            In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
            $endgroup$
            – Henry
            Sep 10 '11 at 22:09












          • $begingroup$
            Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 22:09














          3












          3








          3





          $begingroup$

          You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.



          If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.



          So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.



          Add up the last three (and change to lowest terms) and you have your solution.






          share|cite|improve this answer









          $endgroup$



          You are going to have to work out the probabilities of getting the keyword exactly $n$ times, and either add up the probabilities for $n=2$, $3$ or $4$, or subtract from $1$ the probabilities for $n=1$ or $0$.



          If $N$ is the number of dictionaries where the keyword is found then $$Pr(N=n) = frac{{5 choose n}{4 choose 4-n}}{9 choose 4}$$ since you have to choose $n$ dictionaries from $5$ with the keyword and $4-n$ from $4$ without, choosing $4$ from $9$ overall.



          So for $n=0,1,2,3,4$ this gives probabilities $frac{1}{126},frac{20}{126},frac{60}{126},frac{40}{126},frac{5}{126}$.



          Add up the last three (and change to lowest terms) and you have your solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 '11 at 17:48









          HenryHenry

          99k478164




          99k478164












          • $begingroup$
            @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
            $endgroup$
            – Henry
            Sep 10 '11 at 21:23












          • $begingroup$
            Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:48












          • $begingroup$
            So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:53










          • $begingroup$
            In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
            $endgroup$
            – Henry
            Sep 10 '11 at 22:09












          • $begingroup$
            Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 22:09


















          • $begingroup$
            @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
            $endgroup$
            – Henry
            Sep 10 '11 at 21:23












          • $begingroup$
            Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:48












          • $begingroup$
            So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 21:53










          • $begingroup$
            In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
            $endgroup$
            – Henry
            Sep 10 '11 at 22:09












          • $begingroup$
            Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 22:09
















          $begingroup$
          @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
          $endgroup$
          – Henry
          Sep 10 '11 at 21:23






          $begingroup$
          @Mike: I don't know: the value is the same but the words are not. You have ${5 choose 3}$ where I have ${5 choose 2}$, and you say "ways that 3 of the last 5 databases can contain the keyword" while I say "you have to choose 2 dictionaries from 5 with the keyword". Similarly you have ${9 choose 5}$ where I have ${9 choose 4}$, and you say "ways that 5 of the 9 total databases to contain the keyword" while I say "choosing 4 from 9 overall".
          $endgroup$
          – Henry
          Sep 10 '11 at 21:23














          $begingroup$
          Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 21:48






          $begingroup$
          Suppose there were only 2 databases (rather than 5) that contain the keyword. If I'm following your logic correctly, your answer to that modified problem would be $P(X = 2) = binom{2}{2} binom{7}{2}/binom{9}{4}$ (2 from 2 with keyword, 2 from 7 without keyword, and 4 from 9 overall). The answer, though, is $binom{4}{2}/binom{9}{2}$ (the number of ways to distribute two keywords among a set of 4 divided by the number of ways to distribute two keywords among a set of 9). So I'm not sure that your logic works in general. (Or am I missing something?)
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 21:48














          $begingroup$
          So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 21:53




          $begingroup$
          So I guess it's not the same as my solution after all. (Deleting earlier comment to that effect.)
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 21:53












          $begingroup$
          In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
          $endgroup$
          – Henry
          Sep 10 '11 at 22:09






          $begingroup$
          In general, if there are $d$ dictionaries and $k$ of them have keywords, and you choose $m$ of them then $Pr(N=n) = dfrac{{k choose n}{d-k choose m-n}}{d choose n}$.
          $endgroup$
          – Henry
          Sep 10 '11 at 22:09














          $begingroup$
          Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 22:09




          $begingroup$
          Huh. Even in the example I sort of picked at random, our logic still gives the same numerical answer! Perhaps our answers are equivalent after all, even when they don't look like it. Unfortunately, I don't have time right now to pursue this any further. +1 from me.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 22:09











          3












          $begingroup$

          Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.



          $$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          But



          $$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
          $$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          Thus the two answers give the same result.




          (Original answer.)



          I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.



          $$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.



          I'll let you calculate $P(X = 3)$ and $P(X = 4)$.



          (FYI, $X$ has what's called a hypergeometric distribution.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
            $endgroup$
            – Kelly
            Sep 10 '11 at 16:56










          • $begingroup$
            @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 17:05










          • $begingroup$
            I feel that there is something wrong above! Are we choosing 4 items and not 5?!
            $endgroup$
            – Kelly
            Sep 10 '11 at 17:06










          • $begingroup$
            @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 20:31
















          3












          $begingroup$

          Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.



          $$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          But



          $$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
          $$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          Thus the two answers give the same result.




          (Original answer.)



          I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.



          $$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.



          I'll let you calculate $P(X = 3)$ and $P(X = 4)$.



          (FYI, $X$ has what's called a hypergeometric distribution.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
            $endgroup$
            – Kelly
            Sep 10 '11 at 16:56










          • $begingroup$
            @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 17:05










          • $begingroup$
            I feel that there is something wrong above! Are we choosing 4 items and not 5?!
            $endgroup$
            – Kelly
            Sep 10 '11 at 17:06










          • $begingroup$
            @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 20:31














          3












          3








          3





          $begingroup$

          Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.



          $$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          But



          $$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
          $$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          Thus the two answers give the same result.




          (Original answer.)



          I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.



          $$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.



          I'll let you calculate $P(X = 3)$ and $P(X = 4)$.



          (FYI, $X$ has what's called a hypergeometric distribution.)






          share|cite|improve this answer











          $endgroup$



          Added: Henry's and my answers, while the approaches are different, actually give the same result! To see why, let's do the general case with $N$ databases, $M$ of which contain the keyword, and us choosing $K$ databases. Let $X$ be the number of the $K$ databases chosen that contain the keyword. For $P(X = x)$, Henry's and my logic give the following answers.



          $$text{ Henry: } P(X = x) = frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}}; text{ me: } P(X = x) = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          But



          $$frac{binom{M}{x}binom{N-M}{K-x}}{binom{N}{K}} = frac{M!}{x! (M-x)!} frac{(N-M)!}{(K-x)! (N-M-K+x)!} frac{K! (N-K)!}{N!}$$
          $$= frac{K!}{x! (K-x)!} frac{(N-K)!}{(M-x)! (N-K-M+x)!} frac{M! (N-M)!}{N!} = frac{binom{K}{x} binom{N-K}{M-x}}{binom{N}{M}}.$$



          Thus the two answers give the same result.




          (Original answer.)



          I would approach it differently. Let $X$ be the number of the first four databases in which the keyword is found. You want $P(X geq 2)$.



          $$P(X = 2) = frac{binom{4}{2} binom{5}{3}}{binom{9}{5}}$$ because there are $binom{4}{2}$ ways that 2 of the first 4 databases can contain the keyword, $binom{5}{3}$ ways that 3 of the last 5 databases can contain the keyword, and $binom{9}{5}$ ways that 5 of the 9 total databases to contain the keyword.



          I'll let you calculate $P(X = 3)$ and $P(X = 4)$.



          (FYI, $X$ has what's called a hypergeometric distribution.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 10 '11 at 23:37

























          answered Sep 10 '11 at 16:53









          Mike SpiveyMike Spivey

          42.4k8141232




          42.4k8141232












          • $begingroup$
            So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
            $endgroup$
            – Kelly
            Sep 10 '11 at 16:56










          • $begingroup$
            @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 17:05










          • $begingroup$
            I feel that there is something wrong above! Are we choosing 4 items and not 5?!
            $endgroup$
            – Kelly
            Sep 10 '11 at 17:06










          • $begingroup$
            @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 20:31


















          • $begingroup$
            So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
            $endgroup$
            – Kelly
            Sep 10 '11 at 16:56










          • $begingroup$
            @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 17:05










          • $begingroup$
            I feel that there is something wrong above! Are we choosing 4 items and not 5?!
            $endgroup$
            – Kelly
            Sep 10 '11 at 17:06










          • $begingroup$
            @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
            $endgroup$
            – Mike Spivey
            Sep 10 '11 at 20:31
















          $begingroup$
          So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
          $endgroup$
          – Kelly
          Sep 10 '11 at 16:56




          $begingroup$
          So, in general: $P(X=k)=frac{binom{4}{k} binom{5}{5-k}}{binom{9}{5}} $
          $endgroup$
          – Kelly
          Sep 10 '11 at 16:56












          $begingroup$
          @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 17:05




          $begingroup$
          @Jess: Yep. That's the form of the probability mass function for a hypergeometric distribution.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 17:05












          $begingroup$
          I feel that there is something wrong above! Are we choosing 4 items and not 5?!
          $endgroup$
          – Kelly
          Sep 10 '11 at 17:06




          $begingroup$
          I feel that there is something wrong above! Are we choosing 4 items and not 5?!
          $endgroup$
          – Kelly
          Sep 10 '11 at 17:06












          $begingroup$
          @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 20:31




          $begingroup$
          @Jess: I was looking at the problem from the point of view of placing 5 keywords into 9 databases. So we are choosing 5 items (databases) in which to place the keywords.
          $endgroup$
          – Mike Spivey
          Sep 10 '11 at 20:31











          0












          $begingroup$

          Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.



          Let $c=4$ databases are chosen and $s$ databases have the keyword.



          The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.



          There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.



          Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.



          Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.



          Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.



          Hence, the required probability is:
          $$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
          frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
          frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$

          Note that you can also use the complement, i.e.:
          $$P(sge 2)=1-P(s=0)-P(s=1).$$



          Thus, in general, the formula is:
          $$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$



          Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Useful, after the answers from 7 years ago?
            $endgroup$
            – Did
            Jan 1 at 12:12










          • $begingroup$
            @Did, I tried to summarize and generalize the other answers.
            $endgroup$
            – farruhota
            Jan 1 at 12:21










          • $begingroup$
            How did you fall on this question and why did you decide to "reawaken" it?
            $endgroup$
            – Did
            Jan 1 at 12:24










          • $begingroup$
            @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
            $endgroup$
            – farruhota
            Jan 1 at 12:29










          • $begingroup$
            Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
            $endgroup$
            – o0omycomputero0o
            Jan 1 at 15:03
















          0












          $begingroup$

          Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.



          Let $c=4$ databases are chosen and $s$ databases have the keyword.



          The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.



          There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.



          Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.



          Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.



          Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.



          Hence, the required probability is:
          $$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
          frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
          frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$

          Note that you can also use the complement, i.e.:
          $$P(sge 2)=1-P(s=0)-P(s=1).$$



          Thus, in general, the formula is:
          $$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$



          Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Useful, after the answers from 7 years ago?
            $endgroup$
            – Did
            Jan 1 at 12:12










          • $begingroup$
            @Did, I tried to summarize and generalize the other answers.
            $endgroup$
            – farruhota
            Jan 1 at 12:21










          • $begingroup$
            How did you fall on this question and why did you decide to "reawaken" it?
            $endgroup$
            – Did
            Jan 1 at 12:24










          • $begingroup$
            @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
            $endgroup$
            – farruhota
            Jan 1 at 12:29










          • $begingroup$
            Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
            $endgroup$
            – o0omycomputero0o
            Jan 1 at 15:03














          0












          0








          0





          $begingroup$

          Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.



          Let $c=4$ databases are chosen and $s$ databases have the keyword.



          The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.



          There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.



          Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.



          Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.



          Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.



          Hence, the required probability is:
          $$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
          frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
          frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$

          Note that you can also use the complement, i.e.:
          $$P(sge 2)=1-P(s=0)-P(s=1).$$



          Thus, in general, the formula is:
          $$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$



          Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.






          share|cite|improve this answer









          $endgroup$



          Let $t=9$ be total number of databases, of which $n=4$ have no and $k=5$ have the keyword.



          Let $c=4$ databases are chosen and $s$ databases have the keyword.



          The question is asking to find $sge 2$, i.e. $s=color{red}2,color{green}3,color{blue}4$.



          There are ${tchoose c}={9choose 4}$ ways to choose $c$ databases from total $t$ databases.



          Case 1: There are ${kchoose 2}={5choose color{red}2}$ ways to choose from $k$ and ${nchoose 2}={4choose 2}$ ways from $n$, hence: ${5choose 2}{4choose 2}$.



          Case 2: There are ${kchoose 3}={5choose color{green}3}$ ways to choose from $k$ and ${nchoose 1}={4choose 1}$ ways from $n$, hence: ${5choose 3}{4choose 1}$.



          Case 3: There are ${kchoose 4}={5choose color{blue}4}$ ways to choose from $k$ and ${nchoose 0}={4choose 0}$ ways from $n$, hence: ${5choose 4}{4choose 0}$.



          Hence, the required probability is:
          $$P(sge 2)=P(s=2)+P(s=3)+P(s=4)=\
          frac{{5choose 2}{4choose 2}}{9choose 4}+frac{{5choose 3}{4choose 1}}{9choose 4}+frac{{5choose 4}{4choose 0}}{9choose 4}=\
          frac{60+40+5}{126}=frac{35}{42}approx 0.83.$$

          Note that you can also use the complement, i.e.:
          $$P(sge 2)=1-P(s=0)-P(s=1).$$



          Thus, in general, the formula is:
          $$P(sge r)=sum_{i=r}^c P(s=i)=sum_{i=r}^c frac{{kchoose i}{nchoose c-i}}{{tchoose c}}.$$



          Keys: $t$-total, $n$-no keyword, $k$-keyword, $c$-chosen, $s$-success, $r$-minimum required success.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 11:26









          farruhotafarruhota

          19.7k2738




          19.7k2738












          • $begingroup$
            Useful, after the answers from 7 years ago?
            $endgroup$
            – Did
            Jan 1 at 12:12










          • $begingroup$
            @Did, I tried to summarize and generalize the other answers.
            $endgroup$
            – farruhota
            Jan 1 at 12:21










          • $begingroup$
            How did you fall on this question and why did you decide to "reawaken" it?
            $endgroup$
            – Did
            Jan 1 at 12:24










          • $begingroup$
            @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
            $endgroup$
            – farruhota
            Jan 1 at 12:29










          • $begingroup$
            Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
            $endgroup$
            – o0omycomputero0o
            Jan 1 at 15:03


















          • $begingroup$
            Useful, after the answers from 7 years ago?
            $endgroup$
            – Did
            Jan 1 at 12:12










          • $begingroup$
            @Did, I tried to summarize and generalize the other answers.
            $endgroup$
            – farruhota
            Jan 1 at 12:21










          • $begingroup$
            How did you fall on this question and why did you decide to "reawaken" it?
            $endgroup$
            – Did
            Jan 1 at 12:24










          • $begingroup$
            @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
            $endgroup$
            – farruhota
            Jan 1 at 12:29










          • $begingroup$
            Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
            $endgroup$
            – o0omycomputero0o
            Jan 1 at 15:03
















          $begingroup$
          Useful, after the answers from 7 years ago?
          $endgroup$
          – Did
          Jan 1 at 12:12




          $begingroup$
          Useful, after the answers from 7 years ago?
          $endgroup$
          – Did
          Jan 1 at 12:12












          $begingroup$
          @Did, I tried to summarize and generalize the other answers.
          $endgroup$
          – farruhota
          Jan 1 at 12:21




          $begingroup$
          @Did, I tried to summarize and generalize the other answers.
          $endgroup$
          – farruhota
          Jan 1 at 12:21












          $begingroup$
          How did you fall on this question and why did you decide to "reawaken" it?
          $endgroup$
          – Did
          Jan 1 at 12:24




          $begingroup$
          How did you fall on this question and why did you decide to "reawaken" it?
          $endgroup$
          – Did
          Jan 1 at 12:24












          $begingroup$
          @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
          $endgroup$
          – farruhota
          Jan 1 at 12:29




          $begingroup$
          @Did, the third answer brought it up, I read the discussions in comments and decided to organize the answers. Oh well, I see the OP has been off since then. So, it looks the problem will hang out unaccepted.
          $endgroup$
          – farruhota
          Jan 1 at 12:29












          $begingroup$
          Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
          $endgroup$
          – o0omycomputero0o
          Jan 1 at 15:03




          $begingroup$
          Thanks. I've read the question and all answers again; seem that I misunderstood the question (4 searched databases here mean each search hit distinguished database); the answer has been deleted.
          $endgroup$
          – o0omycomputero0o
          Jan 1 at 15:03


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f63356%2fconditional-probability-problem%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅