Question

An die is tossed twice. Find the probability of getting $4,5or6$ on the toss and $1,2,3or4$ on the second toss.

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{1}{3}$

Total possible outcomes $6$

Favorable outcomes for first event $4,5,6$

$\text{Probability of first event}$ $=\frac{3}{6}=\frac{1}{2}$

Favorable outcomes for second event $1,2,3,4$

$\text{Probability of first event}$ $=\frac{4}{6}=\frac{2}{3}$

The events are independant

So $\text{Required proabability}$ $=\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}$