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

Answer 1

Answer: (A)

$\frac { 1 } { 18 }$

Answer 2

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 apearing 4 is B. So we have to find $P \left( \frac { A } { B } \right)$ .

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

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

Now $n ( B ) = 36$ , 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, $n ( A \cap B ) = 2$ , $n ( B ) = 36$; $\therefore$ $P \left( \frac { A } { B } \right) = \frac { 2 } { 36 } = \frac { 1 } { 18 }$ .