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Chapter 2: Q 2.12 (page 54)

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

Expert verified

P=C46脳C24脳2脳3(10)!5!25=.5714

Step by step solution

01

step 1

There are (10)!/25 different divisions of the 10 players into a first roommate pair, a second roommate pair, and so on. Hence, there are (10)!/(5!25) divisions into 5 roommate pairs.

02

Step 2

There are (C46)(C24) 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!22)=3 ways of making two roommate pairs from the remaining four frontcourt players

03

Step 3

The desired probability is =C46脳C24脳2脳3(10!)5!25=0.5714

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