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Expected Value question about married couples

1

$begingroup$

A group of 36 people, consisting of 18 married couples, are put into random teams of three for a scavenger hunt. In particular, let $a_1,a_2,ldots,a_36$ be the people, and assume that for each $iin[1..18]$, the pair $(a_2i-1,a_2i)$ is a married couple. An event organizer randomly permutes the people, and then for each $kin[1..12]$ assigns the people in positions $3k-2$, $3k-1$, and $3k$ to Team $k$. How many of the teams should we expect to contain married couples?

Am I in the right track with the following :

Let $A_i$ be the probability that a given team has a married couple. $P(A_i)$ = $(18times 34)$/$36choose3$

So we can think of $A_i$ = 1 if team has a married couple and $A_i$ = 0 if the team doesn't have a married couple.

So we can find the expected value of A if we find the expected values of $A_1$ + $A_2$ + ....$A_18$.

Since we know the probability that a team has a married couple is $(18times 34)$/$36choose3$, would the expected number of teams with married couples be :

E[A] = E[$A_1$ + $A_2$ + ....$A_18$] = 18 x $(A_i)$ or (18) x $(18times 34)$/$36choose3$

probability

share|cite|improve this question

edited Apr 1 at 2:15

Robin

asked Apr 1 at 0:37

RobinRobin

675

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well...what's the probability that a randomly selected triple has no couples?
  $endgroup$
  – lulu
  Apr 1 at 1:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Suggestion: Try first to study a simpler problem with 6 persons and 2 teams. And then, 12 persons and 4 teams. Is there a pattern?
  $endgroup$
  – Ertxiem
  Apr 1 at 1:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lulu would it be 1 - ($(18times 34)$/$36choose3$)?
  $endgroup$
  – Robin
  Apr 1 at 2:00
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it would! So then you know the probability that a randomly selected triple does contain a married couple. And once you know that linearity tells you...
  $endgroup$
  – lulu
  Apr 1 at 9:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note: Your expression, while correct, is unnecessarily complicated. Having chosen one person, the next does not make a couple with probability $frac 3435$. Then, conditioned on that, the third also fails to make a couple with probability $frac 3234$. Hence the answer is the product $frac 3435times frac 3234$. As I say, that's the same as what you wrote, but maybe it's easier to work with.
  $endgroup$
  – lulu
  Apr 1 at 9:32
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

A group of 36 people, consisting of 18 married couples, are put into random teams of three for a scavenger hunt. In particular, let $a_1,a_2,ldots,a_36$ be the people, and assume that for each $iin[1..18]$, the pair $(a_2i-1,a_2i)$ is a married couple. An event organizer randomly permutes the people, and then for each $kin[1..12]$ assigns the people in positions $3k-2$, $3k-1$, and $3k$ to Team $k$. How many of the teams should we expect to contain married couples?

Am I in the right track with the following :

Let $A_i$ be the probability that a given team has a married couple. $P(A_i)$ = $(18times 34)$/$36choose3$

So we can think of $A_i$ = 1 if team has a married couple and $A_i$ = 0 if the team doesn't have a married couple.

So we can find the expected value of A if we find the expected values of $A_1$ + $A_2$ + ....$A_18$.

Since we know the probability that a team has a married couple is $(18times 34)$/$36choose3$, would the expected number of teams with married couples be :

E[A] = E[$A_1$ + $A_2$ + ....$A_18$] = 18 x $(A_i)$ or (18) x $(18times 34)$/$36choose3$

probability

share|cite|improve this question

edited Apr 1 at 2:15

Robin

asked Apr 1 at 0:37

RobinRobin

675

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well...what's the probability that a randomly selected triple has no couples?
  $endgroup$
  – lulu
  Apr 1 at 1:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Suggestion: Try first to study a simpler problem with 6 persons and 2 teams. And then, 12 persons and 4 teams. Is there a pattern?
  $endgroup$
  – Ertxiem
  Apr 1 at 1:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lulu would it be 1 - ($(18times 34)$/$36choose3$)?
  $endgroup$
  – Robin
  Apr 1 at 2:00
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it would! So then you know the probability that a randomly selected triple does contain a married couple. And once you know that linearity tells you...
  $endgroup$
  – lulu
  Apr 1 at 9:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note: Your expression, while correct, is unnecessarily complicated. Having chosen one person, the next does not make a couple with probability $frac 3435$. Then, conditioned on that, the third also fails to make a couple with probability $frac 3234$. Hence the answer is the product $frac 3435times frac 3234$. As I say, that's the same as what you wrote, but maybe it's easier to work with.
  $endgroup$
  – lulu
  Apr 1 at 9:32
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

A group of 36 people, consisting of 18 married couples, are put into random teams of three for a scavenger hunt. In particular, let $a_1,a_2,ldots,a_36$ be the people, and assume that for each $iin[1..18]$, the pair $(a_2i-1,a_2i)$ is a married couple. An event organizer randomly permutes the people, and then for each $kin[1..12]$ assigns the people in positions $3k-2$, $3k-1$, and $3k$ to Team $k$. How many of the teams should we expect to contain married couples?

Am I in the right track with the following :

Let $A_i$ be the probability that a given team has a married couple. $P(A_i)$ = $(18times 34)$/$36choose3$

So we can think of $A_i$ = 1 if team has a married couple and $A_i$ = 0 if the team doesn't have a married couple.

So we can find the expected value of A if we find the expected values of $A_1$ + $A_2$ + ....$A_18$.

Since we know the probability that a team has a married couple is $(18times 34)$/$36choose3$, would the expected number of teams with married couples be :

E[A] = E[$A_1$ + $A_2$ + ....$A_18$] = 18 x $(A_i)$ or (18) x $(18times 34)$/$36choose3$

probability

share|cite|improve this question

edited Apr 1 at 2:15

Robin

asked Apr 1 at 0:37

RobinRobin

675

$endgroup$

share|cite|improve this question

edited Apr 1 at 2:15

Robin

asked Apr 1 at 0:37

RobinRobin

675

share|cite|improve this question

edited Apr 1 at 2:15

Robin

asked Apr 1 at 0:37

RobinRobin

675

asked Apr 1 at 0:37

RobinRobin

675

asked Apr 1 at 0:37

RobinRobin

675