What is the probability that 3 randomly selected letters of the alphabet match any 3 of 5 letters randomly selected earlier?

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On each occasion (selection of 3, then selection of 5) the same letter of the alphabet cannot be selected more than once.



I am no mathematician but I arrived at the answer 1/260
(5/26 * 4/25 * 3/24)



Not sure this is correct though!



Any help would be appreciated.










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  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Sep 10 at 10:51






  • 2




    Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
    – lulu
    Sep 10 at 11:01










  • My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
    – R Johnson
    Sep 10 at 11:22










  • 65,780 combinations of 5 letters from 26.
    – R Johnson
    Sep 10 at 11:44










  • Not sure how to calculate how many of these contain the choice of 3.
    – R Johnson
    Sep 10 at 11:47














up vote
0
down vote

favorite












On each occasion (selection of 3, then selection of 5) the same letter of the alphabet cannot be selected more than once.



I am no mathematician but I arrived at the answer 1/260
(5/26 * 4/25 * 3/24)



Not sure this is correct though!



Any help would be appreciated.










share|cite|improve this question



















  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Sep 10 at 10:51






  • 2




    Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
    – lulu
    Sep 10 at 11:01










  • My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
    – R Johnson
    Sep 10 at 11:22










  • 65,780 combinations of 5 letters from 26.
    – R Johnson
    Sep 10 at 11:44










  • Not sure how to calculate how many of these contain the choice of 3.
    – R Johnson
    Sep 10 at 11:47












up vote
0
down vote

favorite









up vote
0
down vote

favorite











On each occasion (selection of 3, then selection of 5) the same letter of the alphabet cannot be selected more than once.



I am no mathematician but I arrived at the answer 1/260
(5/26 * 4/25 * 3/24)



Not sure this is correct though!



Any help would be appreciated.










share|cite|improve this question















On each occasion (selection of 3, then selection of 5) the same letter of the alphabet cannot be selected more than once.



I am no mathematician but I arrived at the answer 1/260
(5/26 * 4/25 * 3/24)



Not sure this is correct though!



Any help would be appreciated.







probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 10 at 10:58

























asked Sep 10 at 10:47









R Johnson

33




33







  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Sep 10 at 10:51






  • 2




    Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
    – lulu
    Sep 10 at 11:01










  • My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
    – R Johnson
    Sep 10 at 11:22










  • 65,780 combinations of 5 letters from 26.
    – R Johnson
    Sep 10 at 11:44










  • Not sure how to calculate how many of these contain the choice of 3.
    – R Johnson
    Sep 10 at 11:47












  • 1




    Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
    – José Carlos Santos
    Sep 10 at 10:51






  • 2




    Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
    – lulu
    Sep 10 at 11:01










  • My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
    – R Johnson
    Sep 10 at 11:22










  • 65,780 combinations of 5 letters from 26.
    – R Johnson
    Sep 10 at 11:44










  • Not sure how to calculate how many of these contain the choice of 3.
    – R Johnson
    Sep 10 at 11:47







1




1




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 10 at 10:51




Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 10 at 10:51




2




2




Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
– lulu
Sep 10 at 11:01




Where did your answer come from? Hint: how many $5$ letter draws are possible? Given a choice of $3$, how many $5$ letter draws contain those three?
– lulu
Sep 10 at 11:01












My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
– R Johnson
Sep 10 at 11:22




My approach was that there is a 5 in 26 chance of selecting one of the 5, followed by a 4 in 25 chance etc... but as I said I’m no mathematician lol. Thanks for the hint, I’ll see if I can work that out
– R Johnson
Sep 10 at 11:22












65,780 combinations of 5 letters from 26.
– R Johnson
Sep 10 at 11:44




65,780 combinations of 5 letters from 26.
– R Johnson
Sep 10 at 11:44












Not sure how to calculate how many of these contain the choice of 3.
– R Johnson
Sep 10 at 11:47




Not sure how to calculate how many of these contain the choice of 3.
– R Johnson
Sep 10 at 11:47










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Fix the $5$-letter draw.



There are $binom53=10$ $3$-letter combination that satisify the condition.



There are $binom263$ possible $3$-letter draws.



Therefore the probability is $frac10binom263$, or equivalently, $frac526timesfrac425timesfrac324$






share|cite|improve this answer






















  • @N.F Taussig I was overcomplicating the problem.
    – abc...
    Sep 10 at 12:11










  • Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
    – N. F. Taussig
    Sep 10 at 12:13










  • @N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
    – R Johnson
    Sep 10 at 12:52










  • @RJohnson It is the correct probability. Odds and probability are not synonymous.
    – N. F. Taussig
    Sep 10 at 12:55










  • Thanks. Your help is much appreciated.
    – R Johnson
    Sep 10 at 13:30










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Fix the $5$-letter draw.



There are $binom53=10$ $3$-letter combination that satisify the condition.



There are $binom263$ possible $3$-letter draws.



Therefore the probability is $frac10binom263$, or equivalently, $frac526timesfrac425timesfrac324$






share|cite|improve this answer






















  • @N.F Taussig I was overcomplicating the problem.
    – abc...
    Sep 10 at 12:11










  • Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
    – N. F. Taussig
    Sep 10 at 12:13










  • @N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
    – R Johnson
    Sep 10 at 12:52










  • @RJohnson It is the correct probability. Odds and probability are not synonymous.
    – N. F. Taussig
    Sep 10 at 12:55










  • Thanks. Your help is much appreciated.
    – R Johnson
    Sep 10 at 13:30














up vote
1
down vote



accepted










Fix the $5$-letter draw.



There are $binom53=10$ $3$-letter combination that satisify the condition.



There are $binom263$ possible $3$-letter draws.



Therefore the probability is $frac10binom263$, or equivalently, $frac526timesfrac425timesfrac324$






share|cite|improve this answer






















  • @N.F Taussig I was overcomplicating the problem.
    – abc...
    Sep 10 at 12:11










  • Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
    – N. F. Taussig
    Sep 10 at 12:13










  • @N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
    – R Johnson
    Sep 10 at 12:52










  • @RJohnson It is the correct probability. Odds and probability are not synonymous.
    – N. F. Taussig
    Sep 10 at 12:55










  • Thanks. Your help is much appreciated.
    – R Johnson
    Sep 10 at 13:30












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Fix the $5$-letter draw.



There are $binom53=10$ $3$-letter combination that satisify the condition.



There are $binom263$ possible $3$-letter draws.



Therefore the probability is $frac10binom263$, or equivalently, $frac526timesfrac425timesfrac324$






share|cite|improve this answer














Fix the $5$-letter draw.



There are $binom53=10$ $3$-letter combination that satisify the condition.



There are $binom263$ possible $3$-letter draws.



Therefore the probability is $frac10binom263$, or equivalently, $frac526timesfrac425timesfrac324$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 10 at 22:27

























answered Sep 10 at 12:10









abc...

2,168529




2,168529











  • @N.F Taussig I was overcomplicating the problem.
    – abc...
    Sep 10 at 12:11










  • Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
    – N. F. Taussig
    Sep 10 at 12:13










  • @N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
    – R Johnson
    Sep 10 at 12:52










  • @RJohnson It is the correct probability. Odds and probability are not synonymous.
    – N. F. Taussig
    Sep 10 at 12:55










  • Thanks. Your help is much appreciated.
    – R Johnson
    Sep 10 at 13:30
















  • @N.F Taussig I was overcomplicating the problem.
    – abc...
    Sep 10 at 12:11










  • Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
    – N. F. Taussig
    Sep 10 at 12:13










  • @N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
    – R Johnson
    Sep 10 at 12:52










  • @RJohnson It is the correct probability. Odds and probability are not synonymous.
    – N. F. Taussig
    Sep 10 at 12:55










  • Thanks. Your help is much appreciated.
    – R Johnson
    Sep 10 at 13:30















@N.F Taussig I was overcomplicating the problem.
– abc...
Sep 10 at 12:11




@N.F Taussig I was overcomplicating the problem.
– abc...
Sep 10 at 12:11












Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
– N. F. Taussig
Sep 10 at 12:13




Usually, the binomial coefficient is written $binomnk = fracn!k!(n - k)!$. Otherwise, it looks good.
– N. F. Taussig
Sep 10 at 12:13












@N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
– R Johnson
Sep 10 at 12:52




@N.F.Taussig thanks! Just to clarify, is 1 in 260 the correct odds of this outcome (sorry about terminology!)
– R Johnson
Sep 10 at 12:52












@RJohnson It is the correct probability. Odds and probability are not synonymous.
– N. F. Taussig
Sep 10 at 12:55




@RJohnson It is the correct probability. Odds and probability are not synonymous.
– N. F. Taussig
Sep 10 at 12:55












Thanks. Your help is much appreciated.
– R Johnson
Sep 10 at 13:30




Thanks. Your help is much appreciated.
– R Johnson
Sep 10 at 13:30

















 

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