Q. 2.31 - A First Course in Probability

Question

A $3$-personal basketball team consists of a guard, a forward, and a center.

$\left(a\right)$If a person is chosen at random from each of three different such teams, what is the probability of selecting a complete team?

$\left(b\right)$What is the probability that all $3$players selected play the same position?

Step-by-Step Solution

Answer

$$\begin{gathered} a\left)\frac{2}{9} \\ b\right)\frac{1}{9} \end{gathered}$$

Step 1 Given Information.

A $3$personal basketball team consists of a guard, a forward, and a center.

Step 2 Part (a) Explanation.

We have$3$ choices for a guard from the$3$ teams.$2$ choices for a forward once we have chosen the guard and no choice for a center once we have chosen a guard and a forward. This gives us $3\times 2=6$possible teams which have each member from a different team.  Total possible teams will simply be$27$ as each member can be chosen from $3$different teams having $3$players each$\left(3^3\right)$.

Step 3 Part (b) Explanation.

The probability of the team being all $3$guards is$\frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}=\frac{1}{27}$.

Same probability for the team being all forwards and all centers.  This is calculated by using the fact that choosing a guard/forward/center from a team has a probability$\frac{1}{3}$.