Question

A dies is thrown three times and the sum of three numbers obtained is $15$. The probability of first throw being $5$ is:

• A. $\frac{3}{10}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{3}{10}$

A die is thrown $3$ times, so total probability $6^3=216$

$\Rightarrow$ Sum of die in three throws $=15$

There are three cares by which sum $15$ is obtained,

$1.)5,6,4$

$2).5,5,5$

$3).6,6,3$

$\Rightarrow$ The probability of $5$ appearing on $1st$ throw of die $=\frac{2\times 5}{216}$

and $5$ appears in the $3$ card $=\frac{3}{216}$

$\therefore$ Total probability $=\frac{\frac{3}{216}}{\frac{10}{216}}=\frac{3}{10}$

Hence, the answer is $\frac{3}{10}.$