Question

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

• A. \(\frac25\)
• B. \(\frac15\)
• C. \(\frac16\)
• D. None of these

Answer 1

The Correct Option is B

To find the chance that the first roll was a four given the total sum of rolls is 15, we use conditional probability. Let \( A \) be the event that the first roll is 4, and let \( B \) be the event that the total sum is 15.
The probability we're looking for is \( P(A|B) \), the probability of \( A \) given \( B \). This is calculated as \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
The total number of outcomes for rolling a dice three times is \( 6^3 = 216 \).
Let’s calculate \( P(B) \):
Consider all combinations that add up to 15. Using generating functions or enumeration:
Possible triplets (permutations):
\( (6,6,3), (6,5,4), (6,4,5), (6,3,6), (5,6,4), (5,5,5), (5,4,6), (4,6,5), (4,5,6), (4,4,7) \).
After correcting sums beyond feasible dice face values, valid combinations: \( (6,6,3), (6,5,4), (5,5,5), \) each warranted permuting non-unique values. Count = 10.
\( P(B) = \frac{10}{216} \).
Now calculate \( P(A \cap B) \): If the first roll is 4, solve \( x+y = 11 \) with outcomes: \( (6,5), (5,6), (4,7) \) where dice value cap renders (4,7) unusable. Count: 2.
Thus, \( P(A \cap B) = \frac{2}{216} \).
Substitute into conditional formula:
\( P(A|B) = \frac{\frac{2}{216}}{\frac{10}{216}} = \frac{2}{10} = \frac{1}{5} \).
Therefore, the chance that the first roll was a four is \(\frac{1}{5}\).