Q 2.12 A basketball team consists of 6 ... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

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= $\frac{{}^{6}C_{4} 脳 {}^{4}C_{2} 脳 2 脳 3}{\frac{(10)!}{5! 2^{5}}} =$ .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 = $\frac{{}^{6}C_{4} 脳 {}^{4}C_{2} 脳 2 脳 3}{\frac{(10!)}{5! 2^{5}}} = 0.5714$

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91影视

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

Step by step solution

step 1

Step 2

Step 3

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