Chapter 13, Problem 41SGA

To determine

To find:The required probability.

Answer to Problem 41SGA

The required probability is $\frac{2}{5}$ .

Explanation of Solution

Given:

Two numbers cubes are tossed.

Calculation:

The probability is given as,

  $P\left(E\right)=\frac{s}{n}$

Here, s is the favorable events and n is the total number of outcomes.

Calculate the probability for getting sum of number on the cube is less than 5.

  $P\left(A\right)=\frac{6}{36}$

Calculate the probability for exactly one on each cube.

  $P\left(B\right)=\frac{10}{36}$

Calculate the probability of A and B.

  $P\left(A\text{ and }B\right)=\frac{4}{36}$

The required probability is calculated as,

  $\begin{array}{c}P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}\\ =\frac{\frac{4}{36}}{\frac{10}{36}}\\ =\frac{2}{5}\end{array}$

Thus, the required probability is $\frac{2}{5}$ .