Question

State whether the statement is true/false.

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

• A. True
• B. False

Answer 1

The correct option is A True

Total possible outcomes $6$

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

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

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

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

The events are independant

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