Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{5}{6}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $\frac{1}{3}$

S = {1, 2, 3, 4, 5, 6}, for each case
Let A = event of getting a 1, 2, 3 or 4 on the first toss and B = event of getting a 4, 5 or 6 on the second toss
Then, clearly A and B are independent events.
$\therefore P\left(A\right)=\frac{4}{6}=\frac{2}{3}$ and $P\left(B\right)=\frac{3}{6}=\frac{1}{2}$
So, required probability = $P(A\cap B)$
= $P\left(A\right).P\left(B\right)=\frac{2}{3}.\frac{1}{2}=\frac{1}{3}$