Question

If $12$ identical coins are distributed among three children at random. The probability of distributing so that each child gets atleast two coins is $\frac{7k}{3^{12}}$ then $k$ is

Answer 1

Since, each coin can be distributed to any of three chilldren. Hence total number of ways $=3\times 3\times 3\cdots 12\text{times}=3^{12}$
So, we have $a+b+c=12\cdots \left(i\right)$ where $a,b,c$ are number of coins with each of three children.
Now, formulating equation so that each child gets atleast two coins i.e. $a=a^{\prime }+2,b=b^{\prime }+2,c=c^{\prime }+2$where $a^{\prime }\ge 0,b^{\prime }\ge 0,c^{\prime }\ge 0$ putting in equation $\left(i\right)$
$a^{\prime }+b^{\prime }+c^{\prime }+6=12$
$a^{\prime }+b^{\prime }+c^{\prime }=6$
Now total favorable cases${=}^8{C}_2=28$
Required probability $=\frac{28}{3^{12}}=\frac{7k}{3^{12}}$
$k=4$