Chapter 13, Problem 4STP

To determine

Find the expected number of times C will roll doubles with two number cubes in 90 trials.

Answer to Problem 4STP

J

Explanation of Solution

Formula Used:

If A and B are independent events ,

  $P(A\text{ or }B)=P(A)+P(B)$

  $P(A\text{ and }B)=P(A)\cdot P(B)$

Calculation:

Rolling two dice at a time are both independent events.

The total number of outcomes = $6\cdot 6=36$

We need doubles , so , the favourable number of outcomes = 6

And they are {(1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)}

  $P(\text{getting double})=\frac{\text{favourable outcome}}{\text{total outcome}}=\frac{6}{36}=\frac{1}{6}$

So, in 90 trials , C will roll doubles with two cubes = $90\cdot \frac{1}{6}=15$ times.

Hence , option J is correct.