Question

If a person throw $3$ dice the probability of getting sum of digit exactly $15$ is

• A. $\frac{5}{72}$
• B. $\frac{5}{108}$
• C. $\frac{5}{36}$
• D. $\frac{1}{72}$

Answer 1

The correct option is B $\frac{5}{108}$

For $dice$ to sum to $15$, the first $2$ need to sum between $a$ & $12$as

$\Rightarrow 12$ is maximum sum &

$\Rightarrow 6$ is maximum on $3^{rd}$ dice

so, for the sum to be $15$

$(15-6)$ a should be maximum sum of first $2$ dice

$\begin{array}{c}sumtobe12\to 1way\\ sumtobe11\to 2ways\\ 10\to 3ways\\ 9\to 4ways\end{array}\}$ $10ways$

For first $2$ dice sum of $ato12$ there exist only one number or thired dice so that sum will be $15$

Total no. of outcomes $=6^3=216$

probability $=\frac{10}{216}=\frac{5}{108}$