Question

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. 1
• B. $\frac { 1 } { 3 }$
• C. $\frac { 7 } { 36 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 1 } { 3 }$

Answer 2

Let $P ( A )$ and $P ( B )$ be the probability of the events then $P ( A$ and $B ) = P ( A ) \cdot P ( B ) = \frac { 1 } { 2 } \times \frac { 2 } { 3 } = \frac { 1 } { 3 }$.