A die is tossed thrice. Find the probability of getting an odd number atleast once.

The correct answer is $\dfrac{7}{8}$

Let A be the event of getting an odd number on a single throw of die.

∴ $P(A)$ = $\dfrac{3}{6}$ = $\dfrac{1}{2}$ and $P(A')$ = $1 - P(A)$ = $1 - \dfrac{1}{2}$ = $\dfrac{1}{2}$


Let AAA denote the event of getting an odd number in each of the three throws.

∴ Required probability = $P$(atleast one odd number)

= $1 - P$(no odd number)

= $1 - P(A'A'A')$ = $1 - P(A') P(A') P(A')$

= $1 - \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$ = $1 - \dfrac{1}{8}$ = $\dfrac{7}{8}$