Question

A die is thrown three times and the sum of three numbers obtained is $15$. The probability of first throw being $4$, is

• A. $\frac{1}{18}$
• B. $\frac{1}{5}$
• C. $\frac{4}{5}$
• D. $\frac{17}{18}$

Answer 1

Answer: (B)

If first throw is four, then sum of numbers appearing on last two throws must be equal to eleven. That means last two throws are $(6,5)$ or $(5,6)$
There is a set having sum of numbers as $15$ is
$\{(3,6,6),(6,3,6),(6,6,3),(4,6,5),(4,5,6),(6,5,4)$,
$(6,4,5),(5,4,6),(5,6,4),(5,5,5)\}=10$.
Now, there are $10$ ways to get the sum as $15$.
$\Rightarrow$ Required probability $=\frac{2}{10}$
$=\frac{1}{5}$.