Question

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

• A. 1/18
• B. 1/36
• C. 1/9
• D. 1/3

Answer 1

The correct option is A

1/18

We have to find the bounded probability to get sum 15 when 4 appears first. Let the event of getting sum 15 of three thrown number is A and the event of appearing 4 is B. So we have to find $P\left(\frac{A}{B}\right)$.

But $P\left(\frac{A}{B}\right)=\frac{n(A\cap B)}{n\left(B\right)}$

When $n(A\cap B)$ and $n\left(B\right)$ respectively denote the number of digits in $A\cap B$ and B.

Now $n\left(B\right)=36$, because first throw is of 4. So another two throws stop by $6\times 6=36$ types. Three dices have only two throws, which starts from 4 and give sum 15 i.e., (4,5,6) and (4,6,5).

So, $n(A\cap B)=2$, $n\left(B\right)=36$;

$\therefore$ $P\left(\frac{A}{B}\right)=\frac{2}{36}=\frac{1}{18}$.