Question

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

A. \(\frac{1}{3}\)

B. \(\frac{2}{3}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{6}\)

Answer 1

Correct answer is A.

A die is tossed twice,

The probability of getting a 4, 5 or 6 in the first trial is 3/6 = P(A)

The probability of getting a 1, 2, 3 or 4 in the second trial is 4/6 =P(B)

As the events are independent, the probability of these two events together will be, P(A).P(B) = 1/3.