Question

An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3 or 4 on the second toss.

Answer 1

$$\begin{gathered} P\left(A\right)=P\left(\text{4, 5 or 6 on first toss}\right)=\frac{18}{36}=\frac{1}{2} \\ P\left(B\right)=P\left(\text{1, 2, 3 or 4 on second toss}\right)=\frac{24}{36}=\frac{2}{3} \\ \text{It is clear that A and B are independent events.} \\ \Rightarrow P\left(A\cap B\right)=P\left(A\right)P\left(B\right) \\ \Rightarrow P\left(A\cap B\right)=\frac{1}{2}\times \frac{2}{3} \\ \therefore P\left(A\cap B\right)=\frac{1}{3} \end{gathered}$$