Question

One die is thrown three times and the sum of the thrown numbers is $15$. The probability for which number $4$ appears in first throw

• A. $\frac{1}{18}$
• B. $\frac{1}{36}$
• C. $\frac{1}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A

$\frac{1}{18}$

Explanation for the correct option:

Given: One die is thrown three times and the sum of the thrown numbers is $15$.

Let $A$ be the event of getting a sum of $15$ of the three thrown numbers and $B$ be the event of getting $4$ on the first thrown

Now, only $2$ cases are possible when sum of numbers appearing on the dice is $15$ and first number is $4$ and they are $\left(4,6,5\right)$ and $\left(4,5,6\right)$

and possible cases when first thrown is $4$ will be $36$ as other two thrown have total possible cases $6\times 6$

Thus, the required probability is:

$$\begin{gathered} P\left(\frac{A}{B}\right)=\frac{n(A\cap B)}{n\left(B\right)} \\ =\frac{2}{36}\mathbf{}\left[\because n(A\cap B)=2beacusethereareonlytwocases\right] \\ =\frac{1}{18}. \end{gathered}$$

Hence the correct answer is option(A).