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Question

Mathematics Question on Conditional Probability

A dice is thrown twice. The probability of getting 4, 5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw is

A

44564

B

44595

C

44563

D

44565

Answer

44564

Explanation

Solution

Let P (A) and P (B) be the probability of the events then P (A and B) = P(A).P(B) =12×23=13 = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}