Question

A basketball team consists of 6 frontcourt and 4 backcourt players. If players are divided into roommates at random,what is the probability that there will be exactly two roommate pairs made up of backcourt and a frontcourt player?

Short answer

P=$\frac{{}^{6}C_4\times {}^{4}C_2\times 2\times 3}{{\displaystyle \frac{\left(10\right)!}{5!2^5}}}=$.5714

Step-by-step solution

step 1

There are (10)!/$2^5$ different divisions of the 10 players into a first roommate pair, a second roommate pair, and so on. Hence, there are (10)!/(5!$2^5$) divisions into 5 roommate pairs.

Step 2

There are (${}^{6}C_4$)(${}^{4}C_2$) ways of choosing the frontcourt and backcourt players to be in the mixed roommate pairs, and then 2 ways of pairing them up.As there is then 1 way to pair up the remaining two backcourt, and 4!/(2!$2^2$)=3 ways of making two roommate pairs from the remaining four frontcourt players

Step 3

The desired probability is =$\frac{{}^{6}C_4\times {}^{4}C_2\times 2\times 3}{{\displaystyle \frac{(10!)}{5!2^5}}}=0.5714$