If you flip three fair coins, what is the probability that you'll get two tails and one head in any order?
$begingroup$
I can't find a solution that doesn't involve listing out all the possible combinations. Is there a way that I can use combinations/permutations to easily calculate this by hand?
For example, when I'm solving "If you flip three fair coins, what is the probability that you'll get exactly two tails?" I can use combinations to find how many ways of getting exactly $2$ tails: $(frac{3!}{2!times1!})$. Then I can divide the answer by $8$, which is the total number of possible combinations $(2^3)$.
probability combinations
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
$endgroup$
add a comment |
$begingroup$
I can't find a solution that doesn't involve listing out all the possible combinations. Is there a way that I can use combinations/permutations to easily calculate this by hand?
For example, when I'm solving "If you flip three fair coins, what is the probability that you'll get exactly two tails?" I can use combinations to find how many ways of getting exactly $2$ tails: $(frac{3!}{2!times1!})$. Then I can divide the answer by $8$, which is the total number of possible combinations $(2^3)$.
probability combinations
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
$endgroup$
add a comment |
$begingroup$
I can't find a solution that doesn't involve listing out all the possible combinations. Is there a way that I can use combinations/permutations to easily calculate this by hand?
For example, when I'm solving "If you flip three fair coins, what is the probability that you'll get exactly two tails?" I can use combinations to find how many ways of getting exactly $2$ tails: $(frac{3!}{2!times1!})$. Then I can divide the answer by $8$, which is the total number of possible combinations $(2^3)$.
probability combinations
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
$endgroup$
I can't find a solution that doesn't involve listing out all the possible combinations. Is there a way that I can use combinations/permutations to easily calculate this by hand?
For example, when I'm solving "If you flip three fair coins, what is the probability that you'll get exactly two tails?" I can use combinations to find how many ways of getting exactly $2$ tails: $(frac{3!}{2!times1!})$. Then I can divide the answer by $8$, which is the total number of possible combinations $(2^3)$.
probability combinations
probability combinations
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
edited Dec 31 '18 at 19:59
Nilkantha Ghosal
416
416
asked Dec 31 '18 at 19:44
carolinecaroline
32
asked Dec 31 '18 at 19:44
carolinecaroline
32
asked Dec 31 '18 at 19:44
carolinecaroline
32
32
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
It's just $$binom{3}{2}*dfrac{1}{2^3}=dfrac{3}{8}$$
Think of it this way.
You have 3 slots.
$text{_ _ _}$
I want two of them to be tail out of 3 (hence the $binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.
Similarly I can choose 1 head out of 3, so that would make it $binom{3}{1}$, and nothing else would change.
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $frac{3}{8}$.
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It's just $$binom{3}{2}*dfrac{1}{2^3}=dfrac{3}{8}$$
Think of it this way.
You have 3 slots.
$text{_ _ _}$
I want two of them to be tail out of 3 (hence the $binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.
Similarly I can choose 1 head out of 3, so that would make it $binom{3}{1}$, and nothing else would change.
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
It's just $$binom{3}{2}*dfrac{1}{2^3}=dfrac{3}{8}$$
Think of it this way.
You have 3 slots.
$text{_ _ _}$
I want two of them to be tail out of 3 (hence the $binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.
Similarly I can choose 1 head out of 3, so that would make it $binom{3}{1}$, and nothing else would change.
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
It's just $$binom{3}{2}*dfrac{1}{2^3}=dfrac{3}{8}$$
Think of it this way.
You have 3 slots.
$text{_ _ _}$
I want two of them to be tail out of 3 (hence the $binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.
Similarly I can choose 1 head out of 3, so that would make it $binom{3}{1}$, and nothing else would change.
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
$endgroup$
It's just $$binom{3}{2}*dfrac{1}{2^3}=dfrac{3}{8}$$
Think of it this way.
You have 3 slots.
$text{_ _ _}$
I want two of them to be tail out of 3 (hence the $binom{3}{2}$), and the probability of $2$ tails and $1$ head is basically $dfrac{1}{2^3}$, as there are $1/2$ probability of head/tail for each toss, and there are 3 tosses.
Similarly I can choose 1 head out of 3, so that would make it $binom{3}{1}$, and nothing else would change.
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
answered Dec 31 '18 at 20:09
K Split XK Split X
4,19611131
4,19611131
add a comment |
add a comment |
$begingroup$
When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $frac{3}{8}$.
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
add a comment |
$begingroup$
When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $frac{3}{8}$.
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
add a comment |
$begingroup$
When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $frac{3}{8}$.
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
$endgroup$
When you flip $3$ coins, you either get a head or a tail. So getting 2 tails and 1 heads is exactly the same event as getting exactly two tails. Thus, as you noted in the last part of your question, the answer is $frac{3}{8}$.
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
edited Dec 31 '18 at 21:06
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
answered Dec 31 '18 at 20:02
Nilkantha GhosalNilkantha Ghosal
416
416
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
add a comment |
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
Ok, no need to be rude. So the "in any order" portion makes absolutely no difference, correct?
$endgroup$
– caroline
Dec 31 '18 at 20:40
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
@caroline In any order simply means that you can get the one head in the first, second or third toss which is what you counted in the second part of your question. Did not mean to be rude. Sorry. :P
$endgroup$
– Nilkantha Ghosal
Dec 31 '18 at 21:09
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
$begingroup$
Ok thanks! That makes a lot of sense. (And no worries, it's hard to tell tone over the internet)
$endgroup$
– caroline
Dec 31 '18 at 23:41
add a comment |
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