Probability proof. Check my answer?












0












$begingroup$


Let $A$ and $B$ be events defined on a sample space $Omega$. Prove that



$$textbf{P}(A cap B) geq 1 - textbf{P}(A^{c})- textbf{P}(B^{c}) $$



Answer:



begin{align*}
textbf{P}(Acap B) & = textbf{P}(A)textbf{P}(B) \
& = (1 - textbf{P}(A^{c}))(1 - textbf{P}(B^{c})) \
& = 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B^{c}) geq 1 - textbf{P}(A^{c}) - textbf{P}(B^{c})
end{align*}






probability proof-verification







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$endgroup$








  • 2




    $begingroup$
    $P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
    $endgroup$
    – ajotatxe
    Jan 13 at 20:58










  • $begingroup$
    Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
    $endgroup$
    – LoveTooNap29
    Jan 13 at 21:04












  • $begingroup$
    You have assumed that $A$ and $B$ are independent events, but there is no need to.
    $endgroup$
    – user1337
    Jan 14 at 21:12
















0












$begingroup$


Let $A$ and $B$ be events defined on a sample space $Omega$. Prove that



$$textbf{P}(A cap B) geq 1 - textbf{P}(A^{c})- textbf{P}(B^{c}) $$



Answer:



begin{align*}
textbf{P}(Acap B) & = textbf{P}(A)textbf{P}(B) \
& = (1 - textbf{P}(A^{c}))(1 - textbf{P}(B^{c})) \
& = 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B^{c}) geq 1 - textbf{P}(A^{c}) - textbf{P}(B^{c})
end{align*}






probability proof-verification







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
    $endgroup$
    – ajotatxe
    Jan 13 at 20:58










  • $begingroup$
    Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
    $endgroup$
    – LoveTooNap29
    Jan 13 at 21:04












  • $begingroup$
    You have assumed that $A$ and $B$ are independent events, but there is no need to.
    $endgroup$
    – user1337
    Jan 14 at 21:12














0












0








0





$begingroup$


Let $A$ and $B$ be events defined on a sample space $Omega$. Prove that



$$textbf{P}(A cap B) geq 1 - textbf{P}(A^{c})- textbf{P}(B^{c}) $$



Answer:



begin{align*}
textbf{P}(Acap B) & = textbf{P}(A)textbf{P}(B) \
& = (1 - textbf{P}(A^{c}))(1 - textbf{P}(B^{c})) \
& = 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B^{c}) geq 1 - textbf{P}(A^{c}) - textbf{P}(B^{c})
end{align*}






probability proof-verification







share|cite|improve this question











$endgroup$




Let $A$ and $B$ be events defined on a sample space $Omega$. Prove that



$$textbf{P}(A cap B) geq 1 - textbf{P}(A^{c})- textbf{P}(B^{c}) $$



Answer:



begin{align*}
textbf{P}(Acap B) & = textbf{P}(A)textbf{P}(B) \
& = (1 - textbf{P}(A^{c}))(1 - textbf{P}(B^{c})) \
& = 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) + textbf{P}(A^{c})textbf{P}(B^{c}) geq 1 - textbf{P}(A^{c}) - textbf{P}(B^{c})
end{align*}






probability proof-verification




probability proof-verification






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share|cite|improve this question













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share|cite|improve this question








edited Jan 13 at 22:50









user1337

46110




46110










asked Jan 13 at 20:57









oamyoamy

1




1








  • 2




    $begingroup$
    $P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
    $endgroup$
    – ajotatxe
    Jan 13 at 20:58










  • $begingroup$
    Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
    $endgroup$
    – LoveTooNap29
    Jan 13 at 21:04












  • $begingroup$
    You have assumed that $A$ and $B$ are independent events, but there is no need to.
    $endgroup$
    – user1337
    Jan 14 at 21:12














  • 2




    $begingroup$
    $P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
    $endgroup$
    – ajotatxe
    Jan 13 at 20:58










  • $begingroup$
    Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
    $endgroup$
    – LoveTooNap29
    Jan 13 at 21:04












  • $begingroup$
    You have assumed that $A$ and $B$ are independent events, but there is no need to.
    $endgroup$
    – user1337
    Jan 14 at 21:12








2




2




$begingroup$
$P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
$endgroup$
– ajotatxe
Jan 13 at 20:58




$begingroup$
$P(A)$ and $P(B)$ are numbers. How do you define $P(A)cap P(B)$?
$endgroup$
– ajotatxe
Jan 13 at 20:58












$begingroup$
Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
$endgroup$
– LoveTooNap29
Jan 13 at 21:04






$begingroup$
Indeed as user ajotatxe remarks: already in the first line, the expression “$P(A)cap P(B)$” is meaningless since $P(cdot)$ is a number and $cap$ is a set operation.
$endgroup$
– LoveTooNap29
Jan 13 at 21:04














$begingroup$
You have assumed that $A$ and $B$ are independent events, but there is no need to.
$endgroup$
– user1337
Jan 14 at 21:12




$begingroup$
You have assumed that $A$ and $B$ are independent events, but there is no need to.
$endgroup$
– user1337
Jan 14 at 21:12










3 Answers
3






active

oldest

votes


















1












$begingroup$

Notice that



begin{align*}
&textbf{P}(Acup B) leq 1 Leftrightarrow textbf{P}(A) + textbf{P}(B) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
& 1 - textbf{P}(A^{c}) + 1 - textbf{P}(B^{c}) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
& 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) leq textbf{P}(Acap B)
end{align*}



Where we have used the following properties
begin{cases}
textbf{P}(X) leq 1\\
textbf{P}(X) = 1 - textbf{P}(X^{c})\\
textbf{P}(Xcup Y) = textbf{P}(X) + textbf{P}(Y) - textbf{P}(Xcap Y)
end{cases}



Hope this helps.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that $P(S^c)=1-P(S)$ and $P(Acup B)le 1$ therefore$$P(A)+P(B)-P(Acap B)le 1\P(A)+P(B)-1le P(Acap B)\P(A)+P(B)-1-1+1le P(Acap B)\-P(A^c)-P(B^c)+1le P(Acap B)$$and finally by rearranging the terms $$P(Acap B)ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$Acup B=S$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You wrote:
      $$ textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B) $$
      I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the
      coin comes up heads. Let $B$ be the event the coin comes up heads. This give us:
      begin{align*}
      textbf{P}(A) &= frac{1}{2} \
      textbf{P}(B) &= frac{1}{2} \
      textbf{P}(Acap B) &= frac{1}{2} \
      textbf{P}(A)textbf{P}(B) &= frac{1}{4} \
      end{align*}
      Hence their is a contradiction and your original statement is wrong.



      I hope this helps.



      Bob






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Notice that



        begin{align*}
        &textbf{P}(Acup B) leq 1 Leftrightarrow textbf{P}(A) + textbf{P}(B) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
        & 1 - textbf{P}(A^{c}) + 1 - textbf{P}(B^{c}) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
        & 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) leq textbf{P}(Acap B)
        end{align*}



        Where we have used the following properties
        begin{cases}
        textbf{P}(X) leq 1\\
        textbf{P}(X) = 1 - textbf{P}(X^{c})\\
        textbf{P}(Xcup Y) = textbf{P}(X) + textbf{P}(Y) - textbf{P}(Xcap Y)
        end{cases}



        Hope this helps.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Notice that



          begin{align*}
          &textbf{P}(Acup B) leq 1 Leftrightarrow textbf{P}(A) + textbf{P}(B) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
          & 1 - textbf{P}(A^{c}) + 1 - textbf{P}(B^{c}) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
          & 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) leq textbf{P}(Acap B)
          end{align*}



          Where we have used the following properties
          begin{cases}
          textbf{P}(X) leq 1\\
          textbf{P}(X) = 1 - textbf{P}(X^{c})\\
          textbf{P}(Xcup Y) = textbf{P}(X) + textbf{P}(Y) - textbf{P}(Xcap Y)
          end{cases}



          Hope this helps.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Notice that



            begin{align*}
            &textbf{P}(Acup B) leq 1 Leftrightarrow textbf{P}(A) + textbf{P}(B) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
            & 1 - textbf{P}(A^{c}) + 1 - textbf{P}(B^{c}) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
            & 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) leq textbf{P}(Acap B)
            end{align*}



            Where we have used the following properties
            begin{cases}
            textbf{P}(X) leq 1\\
            textbf{P}(X) = 1 - textbf{P}(X^{c})\\
            textbf{P}(Xcup Y) = textbf{P}(X) + textbf{P}(Y) - textbf{P}(Xcap Y)
            end{cases}



            Hope this helps.






            share|cite|improve this answer









            $endgroup$



            Notice that



            begin{align*}
            &textbf{P}(Acup B) leq 1 Leftrightarrow textbf{P}(A) + textbf{P}(B) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
            & 1 - textbf{P}(A^{c}) + 1 - textbf{P}(B^{c}) - textbf{P}(Acap B) leq 1 Leftrightarrow\\
            & 1 - textbf{P}(A^{c}) - textbf{P}(B^{c}) leq textbf{P}(Acap B)
            end{align*}



            Where we have used the following properties
            begin{cases}
            textbf{P}(X) leq 1\\
            textbf{P}(X) = 1 - textbf{P}(X^{c})\\
            textbf{P}(Xcup Y) = textbf{P}(X) + textbf{P}(Y) - textbf{P}(Xcap Y)
            end{cases}



            Hope this helps.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 13 at 21:04









            user1337user1337

            46110




            46110























                1












                $begingroup$

                Note that $P(S^c)=1-P(S)$ and $P(Acup B)le 1$ therefore$$P(A)+P(B)-P(Acap B)le 1\P(A)+P(B)-1le P(Acap B)\P(A)+P(B)-1-1+1le P(Acap B)\-P(A^c)-P(B^c)+1le P(Acap B)$$and finally by rearranging the terms $$P(Acap B)ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$Acup B=S$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Note that $P(S^c)=1-P(S)$ and $P(Acup B)le 1$ therefore$$P(A)+P(B)-P(Acap B)le 1\P(A)+P(B)-1le P(Acap B)\P(A)+P(B)-1-1+1le P(Acap B)\-P(A^c)-P(B^c)+1le P(Acap B)$$and finally by rearranging the terms $$P(Acap B)ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$Acup B=S$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Note that $P(S^c)=1-P(S)$ and $P(Acup B)le 1$ therefore$$P(A)+P(B)-P(Acap B)le 1\P(A)+P(B)-1le P(Acap B)\P(A)+P(B)-1-1+1le P(Acap B)\-P(A^c)-P(B^c)+1le P(Acap B)$$and finally by rearranging the terms $$P(Acap B)ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$Acup B=S$$






                    share|cite|improve this answer









                    $endgroup$



                    Note that $P(S^c)=1-P(S)$ and $P(Acup B)le 1$ therefore$$P(A)+P(B)-P(Acap B)le 1\P(A)+P(B)-1le P(Acap B)\P(A)+P(B)-1-1+1le P(Acap B)\-P(A^c)-P(B^c)+1le P(Acap B)$$and finally by rearranging the terms $$P(Acap B)ge 1-P(A^c)-P(B^c)$$with the equality if and only if $$Acup B=S$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 13 at 21:41









                    Mostafa AyazMostafa Ayaz

                    17k31039




                    17k31039























                        0












                        $begingroup$

                        You wrote:
                        $$ textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B) $$
                        I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the
                        coin comes up heads. Let $B$ be the event the coin comes up heads. This give us:
                        begin{align*}
                        textbf{P}(A) &= frac{1}{2} \
                        textbf{P}(B) &= frac{1}{2} \
                        textbf{P}(Acap B) &= frac{1}{2} \
                        textbf{P}(A)textbf{P}(B) &= frac{1}{4} \
                        end{align*}
                        Hence their is a contradiction and your original statement is wrong.



                        I hope this helps.



                        Bob






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You wrote:
                          $$ textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B) $$
                          I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the
                          coin comes up heads. Let $B$ be the event the coin comes up heads. This give us:
                          begin{align*}
                          textbf{P}(A) &= frac{1}{2} \
                          textbf{P}(B) &= frac{1}{2} \
                          textbf{P}(Acap B) &= frac{1}{2} \
                          textbf{P}(A)textbf{P}(B) &= frac{1}{4} \
                          end{align*}
                          Hence their is a contradiction and your original statement is wrong.



                          I hope this helps.



                          Bob






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You wrote:
                            $$ textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B) $$
                            I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the
                            coin comes up heads. Let $B$ be the event the coin comes up heads. This give us:
                            begin{align*}
                            textbf{P}(A) &= frac{1}{2} \
                            textbf{P}(B) &= frac{1}{2} \
                            textbf{P}(Acap B) &= frac{1}{2} \
                            textbf{P}(A)textbf{P}(B) &= frac{1}{4} \
                            end{align*}
                            Hence their is a contradiction and your original statement is wrong.



                            I hope this helps.



                            Bob






                            share|cite|improve this answer









                            $endgroup$



                            You wrote:
                            $$ textbf{P}(Acap B) = textbf{P}(A)textbf{P}(B) $$
                            I claim this is wrong. Suppose that we are flipping a fair coin. Let $A$ be the event the
                            coin comes up heads. Let $B$ be the event the coin comes up heads. This give us:
                            begin{align*}
                            textbf{P}(A) &= frac{1}{2} \
                            textbf{P}(B) &= frac{1}{2} \
                            textbf{P}(Acap B) &= frac{1}{2} \
                            textbf{P}(A)textbf{P}(B) &= frac{1}{4} \
                            end{align*}
                            Hence their is a contradiction and your original statement is wrong.



                            I hope this helps.



                            Bob







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 13 at 23:05









                            BobBob

                            938515




                            938515






























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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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