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The chess club of two school consists of 8 and 9 players.4 member from each club are randomly chosen to participate in a competition of two school.chosen player from one team are then randomly paired against chosen players of another team.suppose Rebecca and Elise are in two chess clubs of different school..what is the probability that

1.Rebecca and Elise will be paired

2.Rebecca and Elise will be chosen to play but they will NOT be paired against Each Other
in Probability by (429 points) | 291 views
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(a) for choosing rebcca and elise 7C3/8C4 * 8C3/9C4

and then pairing them up--there will be 16 pairs possible from 4 in one team to 4 in other team,out of whic there is only 1 apir which satisy.

so, 7C3/8C4 * 8C3/9C4 *1/16

(b) for choosing rebecca and elise 7C3/8C4 * 8C3/9C4

and prob of not pairing them up will be 15/16

hence,7C3/8C4 * 8C3/9C4 *15/16

please correct me

0
not sure though..let's wait for others
0

@Akriti I thought the same way. But ans is given something else here https://sites.math.washington.edu/~hoffman/394/394hw3solutions.pdf

2 Answers

+2 votes
Best answer

We have two sets : one is having 8 elements and another one is having 9 elements. (sets representing the schools here)

We are interested in two particular element x and y (for example in place of Rebecca and Elise) 

These two elements $(x,y)$ belong to different sets. Allocate them in any order in those two sets (schools are indistinct, therefore we need not swap $x$,$y$ and simulate the following process again)

After selecting $4$ elements from each set we can map $(x,a,b,c)$ to $(y,p,q,r)$ in $4!$ ways in a one-to-one and onto mapping. These $4!$ ways also include $3!$ ways in which $x$ is always mapped to $y$ and $3*3!$ ways in which $x$ is never mapped to $y$.

(a).

probability that (x,y) pair results at the end  = 

$\begin{align*} 
\left [ \text{probability of x getting selected} \right ] \cdot \left [ \text{probability of y getting selected} \right ] \cdot \left [ \text{probability of getting (x,y) pair} \right ]
\end{align*}$

=

$\begin{align*} 
\left [ \frac{7C3}{8C4} \right ] \cdot \left [ \frac{8C3}{9C4} \right ] \cdot \left [ \frac{3!}{4!} \right ] = \frac{1}{18} \\
\end{align*}$


(b).

probability that (x,y) pair does not result at the end  =

$\begin{align*} 
\left [ \frac{7C3}{8C4} \right ] \cdot \left [ \frac{8C3}{9C4} \right ] \cdot \left [ \frac{3*3!}{4!} \right ] = \frac{1}{6} \\
\end{align*}$

by Veteran (56.9k points)
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for (A) part,i understood your reason for 3!/4!

but whats wrong with 1/16 as there will be only one favourable case for rebecca and elise to be together among 16 possible pairs)
0 votes
1) Selection 4/8*4/9

Pairing against one another 1/4

So  answer 1/18

2) Selection same

    Pairing not against one another 3/4

Answer 3/18

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