Chapter CSB, Problem 4.11E

To determine

Find the probability $P\left(acomedyDVD,thenachildren\text{'}sDVD\right).$

Answer to Problem 4.11E

  $\frac{1}{16}.$

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 comedy DVD is selected.

The possible outcomes are $3.$

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

  $P\left(A\right)=\frac{3}{16}$

Assume $B$ as the event when second children’s DVD is selected.

As there is no replacement, the total outcome becomes $15.$

So, the probability such that the second children’s DVD is selected is,

  $\begin{array}{l}P\left(B\right)=\frac{5}{15}\\ =\frac{1}{3}\end{array}$

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

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

  $\begin{array}{l}P\left(acomedyDVD,thenachildren\text{'}sDVD\right)=P\left(A\right).P\left(B\right)\\ =\frac{3}{16}.\frac{1}{3}\\ =\frac{1}{16}\end{array}$

Hence, $P\left(acomedyDVD,thenachildren\text{'}sDVD\right)is\frac{1}{16}.$