GATE CS 2012 Q33 — Conditional Probability and Bayes Theorem

Let $X_1$ be the outcome of the first roll of the fair six-sided die, and $X_2$ be the outcome of the second roll (if it occurs). Let $S$ be the sum total of the values.

The first roll $X_1$ can result in any value from $1$ to $6$ with equal probability of $\frac{1}{6}$. We analyze the two main cases based on $X_1$:

Case 1: $X_1 \in \{4, 5, 6\}$ (No second roll)

In this case, the sum total is simply $S = X_1$. For $S \ge 6$, the only favorable outcome is $X_1 = 6$.

• $P(X_1 = 6) = \dfrac{1}{6}$

Case 2: $X_1 \in \{1, 2, 3\}$ (Die is rolled a second time)

Here, the sum total is $S = X_1 + X_2$. We need $X_1 + X_2 \ge 6$, which means $X_2 \ge 6 - X_1$.

• If $X_1 = 1$: We need $X_2 \ge 5$. The favorable outcomes for $X_2$ are $\{5, 6\}$ (2 outcomes).
  Probability = $P(X_1 = 1) \times P(X_2 \ge 5) = \dfrac{1}{6} \times \dfrac{2}{6} = \dfrac{2}{36}$
• If $X_1 = 2$: We need $X_2 \ge 4$. The favorable outcomes for $X_2$ are $\{4, 5, 6\}$ (3 outcomes).
  Probability = $P(X_1 = 2) \times P(X_2 \ge 4) = \dfrac{1}{6} \times \dfrac{3}{6} = \dfrac{3}{36}$
• If $X_1 = 3$: We need $X_2 \ge 3$. The favorable outcomes for $X_2$ are $\{3, 4, 5, 6\}$ (4 outcomes).
  Probability = $P(X_1 = 3) \times P(X_2 \ge 3) = \dfrac{1}{6} \times \dfrac{4}{6} = \dfrac{4}{36}$

Total Probability

Summing the probabilities of all mutually exclusive favorable cases:

Thus, the probability that the sum total of values is at least $6$ is $\dfrac{5}{12}$.