Chapter CSB, Problem 4.9E

To determine

Find the probability $P\left(2actionDVDs\right)$ with replacement.

Answer to Problem 4.9E

  $\frac{1}{4}.$

Explanation of Solution

Given information:

On a shelf, there are $8$ action, $3$ comedy and $5$ chidren’s DVDs.

Assume that $2$ DVDs are selected at random.

Calculation:

Total number of outcomes are,

  $8+3+5=16$

Assume $A$ as the event when first action DVD is selected.

The possible outcomes are $8.$

So, the probability such that the first action DVD is selected is,

  $\begin{array}{l}P\left(A\right)=\frac{8}{16}\\ =\frac{1}{2}\end{array}$

Assume $B$ as the event when second action DVD is selected.

That is, after replacement the possible outcomes are $8$ only.

So, the probability such that the second action DVD is selected is,

  $\begin{array}{l}P\left(B\right)=\frac{8}{16}\\ =\frac{1}{2}\end{array}$

Events $AandB$ are independent events as the occurence of $A$ does not affect the occurence of $B.$

Multilpy the probabilities of independent events $AandB$ to obtain the required probability, that is,

  $\begin{array}{l}P\left(2actionDVDs\right)=P\left(A\right).P\left(B\right)\\ =\frac{1}{2}.\frac{1}{2}\\ =\frac{1}{4}\end{array}$

Hence, $P\left(2actionDVDs\right)=\frac{1}{4}.$