Question

A die is tossed thrice.

Find the probability of getting an odd number at least once.

Answer 1

We have,

Number of possible outcomes on a die$=\{1,2,3,4,5,6\}$

Favourable results getting even number$=(2,4,6)=3$

Step 1: Calculate the probability of even number once

$\begin{array}{rcl}\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{even}\mathrm{number}& =& \frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}\\ & =& \frac{3}{6}\\ & & \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}3\\ & =& \frac{1}{2}\end{array}$

Probability of getting even number once $=\frac{1}{2}$

Step 2: Calculate the probability of even number thrice

$\begin{array}{rcl}\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{even}\mathrm{number}\mathrm{thrice}& =& \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\\ & =& \frac{1}{8}\end{array}$

Step 3: Calculate the probability of getting at least one odd number

Therefore, probability of getting at least one odd number$\begin{array}{rcl}& =& 1-\frac{1}{8}\end{array}$

$=\frac{7}{8}$

Hence, probability of getting an odd number at least once is $\frac{7}{8}$