Question

A dice is rolled three times and the sum of the three numbers on the uppermost face is 15. What is the chance that the first roll was a four?

• A. \(\frac{2}{5}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{6}\)
• D. None of these

Hint:

First list all dice combinations that add to 15, then count which ones start with a 4.

Answer 1

The Correct Option is B

Here is another way to reach the same answer, using a substitution trick instead of listing sets by hand.

Each die value $a$, $b$, $c$ lies between 1 and 6. Write $a' = 6 - a$, $b' = 6 - b$, $c' = 6 - c$, so each of $a'$, $b'$, $c'$ lies between 0 and 5. The condition $a + b + c = 15$ becomes $a' + b' + c' = 18 - 15 = 3$.

Since 3 is small, no variable can exceed 5 anyway, so the number of whole number solutions of $a' + b' + c' = 3$ is just $\binom{3+2}{2} = 10$. This matches the 10 arrangements found by direct listing.

Now apply the same substitution to the condition that the first roll is 4: $a = 4$ means $a' = 2$, so $b' + c' = 1$ with $b', c' \geq 0$. That equation has exactly 2 solutions: $(b', c') = (0,1)$ and $(1,0)$.

So the favourable count is 2 out of a total of 10 equally likely outcomes.

Let's summarize:

• Total ways to get sum 15 in three rolls: 10
• Ways with first roll equal to 4: 2
• Required chance: $\frac{2}{10} = \frac{1}{5}$

This confirms option B without listing every triple by hand.