Question

A die is thrown thrice. Find the probability of getting an odd number at least once.

Answer 1

$$\begin{gathered} P\left(\text{getting an odd number in one throw}\right)=\frac{1}{2} \\ \mathrm{Here},\mathrm{g}\text{etting an odd number in three throws refers to 3 independent events.} \\ P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{2} \\ P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-\left[P\left(A\cap B\right)+\left(B\cap C\right)+\left(C\cap A\right)\right]+P\left(A\cap B\cap C\cap \right) \\ =\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\left[\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\right]+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} \\ =\frac{3}{2}-\frac{3}{4}+\frac{1}{8} \\ =\frac{12-6+1}{8} \\ =\frac{7}{8} \end{gathered}$$