Question

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

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

Answer 1

Answer: (B)

$\frac{1}{5}$

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 $(5,6)$.

Now there are $10$ ways to get the sum as $15$.

i.e., $\{(4,5,6),(4,6,5),...,(3,6,6)\}$

$\Rightarrow$ Required probability$=\frac{2}{10}=\frac{1}{5}$