Two dice are rolled and 3 coins are tossed. The probability of getting doublet on the dice and exactly 2 heads on the coins is 

Two dice are rolled and 3 coins are tossed.

$\mathrm{A}= \mathrm{getting} \mathrm{a} \mathrm{doublet} \mathrm{on} \mathrm{the} \mathrm{dice}=\left\{\left(1,1\right),\left(2,2\right),..\left(6,6\right)\right\}$

$P\left(A\right)=\frac{6}{36}=\frac{1}{6}$

$\mathrm{B}= \mathrm{getting} \mathrm{exactly} 2 \mathrm{heads} \mathrm{when} 3 \mathrm{coins} \mathrm{are} \mathrm{tossed}$

$P\left(B\right)=\frac{{ }^3{C}_2}{2^3}=\frac{3}{8}$

A,B are independent events

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)=\frac{1}{6}\times \frac{3}{8}=\frac{1}{16}$