Q. 2.12 - A First Course in Probability

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 a backcourt and a frontcourt player?

Step-by-Step Solution

Answer

$P\left(A\right)=\frac{4}{7}$

Step 1 Given Information.

A basketball team consists of$6$ frontcourt and$4$ backcourt players. If players are divided into roommates at random.

Step 2 Explanation.

The described situation is equivalent to:

Experiment: $10$people are randomly divided into$5$ pairs. Six of the people are different from the remaining four. Call the six people$f$-s and the other four$b-s)$.

Find: the probability that precisely two of the pairs are $f$and $b$pairs.

Outcome space $S$contains every $5$pair combination of these people.

If all events in $S$are considered equally likely, the probability of event $A\subseteq S$is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where $\left|X\right|$denotes the number of elements in$X$.

The number of elements in$S$ :

Choose the firs pair:$\left(\begin{array}{c}10\\ 2\end{array}\right)$ ways.

Choose the second pair:$\left(\begin{array}{l}8\\ 2\end{array}\right)$ ways.

The total number of such choices is$\left(\begin{array}{c}10\\ 2\end{array}\right)·\left(\begin{array}{l}8\\ 2\end{array}\right)·\left(\begin{array}{l}6\\ 2\end{array}\right)·\left(\begin{array}{l}4\\ 2\end{array}\right)=\frac{10\text{ ! }}{2^5}$.

But this way, every combination was counted $5 !$ times, the permuted pairs were counted separately.

$\left|S\right|=\frac{10!}{5!·2^5}$

Step 3 Explanation.

The event$A$ - there are precisely$2$ pairs of one $f$and one$b$.

There are $\left(\begin{array}{l}6\\ 2\end{array}\right)$choices for $f$s that are paired with $b$s.

There are $\left(\begin{array}{l}4\\ 2\end{array}\right)$choices for $b$s that are paired with $f$s.

The two $f_s$and two $b$s can be paired in $2$ways - $f_1{b}_1,f_2{b}_2$or$f_1{b}_2,f_2{b}_1$.

The two remaining$b$s must be paired together.

The four remaining $f \mathrm{s}$can be paired in $\frac{4!}{2!-2^2}$ways (analogous to the number of elements$S$).

All of these choices have the same number of options independently of the others. By the multiplication principle:

$\left|A\right|=\left(\begin{array}{l}6\\ 2\end{array}\right)·\left(\begin{array}{l}4\\ 2\end{array}\right)·2·\frac{4!}{2!·2^2}=\frac{6!·2·4!}{2^3·2!·2^2}=\frac{6!·4!}{2^5}$

Thus:

$P\left(A\right)=\frac{\frac{6!·4!}{2^8}}{\frac{10!}{5!·2^8}}=\frac{6!5!4!}{10!}=\frac{4}{7}$