Question

A dice is thrown twice. The probability of getting $4,5or6$ in the first throw and $1,2,3or4$ in the second throw is -

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

Answer 1

The correct option is A $\frac{1}{3}$

$Let$ $the$ $first$ $throw$ $be$ $event$ $A$ $and$ $second$ $throw$ $be$ $event$ $B.$
$A$ $and$ $B$ $are$ $independent$ $events.$
$\therefore$ $P(A\cap B)=P\left(A\right).P\left(B\right)$ $=$ $\frac{3}{6}\times \frac{4}{6}=\frac{1}{3}$