Chapter 11, Problem 14CR

To determine

To find the probability.

Answer to Problem 14CR

The probability is $\frac{5}{9}$ of getting a number less than $3$ on at least one number cube.

Explanation of Solution

Given:

There are two number cubes.

Formula used:

  $P\left(\text{event}\right)=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$

Calculation:

Total possible outcomes of rolling two number cubes is:

  $\begin{array}{l}\left(1,1\right)\left(1,2\right)\left(1,3\right)\left(1,4\right)\left(1,5\right)\left(1,6\right)\left(2,1\right)\left(2,2\right)\left(2,3\right)\\ \left(2,4\right)\left(2,5\right)\left(2,6\right)\left(3,1\right)\left(3,2\right)\left(3,3\right)\left(3,4\right)\left(3,5\right)\left(3,6\right)\\ \left(4,1\right)\left(4,2\right)\left(4,3\right)\left(4,4\right)\left(4,5\right)\left(4,6\right)\left(5,1\right)\left(5,2\right)\left(5,3\right)\\ \left(5,4\right)\left(5,5\right)\left(5,6\right)\left(6,1\right)\left(6,2\right)\left(6,3\right)\left(6,4\right)\left(6,5\right)\left(6,6\right)\end{array}$

Total number of possible outcomes $=36$ .

Total number of favorable outcomes of getting a number less than $3$ on at least one number

cube $=20$

Therefore,

  $\begin{array}{l}P\left(\text{event}\right)=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}\\ P=\frac{20}{36}=\frac{5}{9}\end{array}$

Hence,

The probability is $\frac{5}{9}$ of getting a number less than $3$ on at least one number cube.