GATE CS 2014 Set 1 Q.58 (CS Q.48): Probability and Statistics

Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is $X / 1296$ . The value of $X$ is ______________.

Your answer

Official answer and solution

Solution

Transform the sum of dice to a complementary variable representing the deficit from maximum score 24.

Let the outcomes of the four dice be $d_1, d_2, d_3, d_4 \in \{1, 2, 3, 4, 5, 6\}$ .
Total number of outcomes = $6^4 = 1296$ .

We want the number of solutions to:

d_1 + d_2 + d_3 + d_4 = 22

The maximum possible sum of four dice is $6 \times 4 = 24$ .
Define $y_i = 6 - d_i$ . Since $1 \le d_i \le 6$ , we have $0 \le y_i \le 5$ .

Rewriting the equation in terms of $y_i$ :

(6 - y_1) + (6 - y_2) + (6 - y_3) + (6 - y_4) = 22

24 - (y_1 + y_2 + y_3 + y_4) = 22

y_1 + y_2 + y_3 + y_4 = 2

Since the sum is only 2, no $y_i$ can exceed 5, so the constraint $y_i \le 5$ is automatically satisfied for all $y_i \ge 0$ .

The number of non-negative integer solutions to $y_1 + y_2 + y_3 + y_4 = 2$ is given by the stars-and-bars formula:

\binom{n + k - 1}{k - 1} = \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10

These 10 combinations correspond to:

One die having value 4, three dice having value 6: $\binom{4}{1} = 4$ ways.
Two dice having value 5, two dice having value 6: $\binom{4}{2} = 6$ ways.
Total favorable outcomes $X = 4 + 6 = 10$ .

Thus, the probability is $10 / 1296$ , so $X = 10$ .

Final answer

Common trap

Listing permutations manually and missing either the $(4, 6, 6, 6)$ or $(5, 5, 6, 6)$ combinations.

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