GATE CS 2012 Q.33 (CS Q.33): Probability and Statistics

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

10/21

5/12

2/3

1/6

Official answer and solution

Official answer: B
From: Official GATE 2012 answer key

Solution

Apply the law of total probability conditioned on the outcome of the first die roll.

Let $X_1$ be the value of the first die roll, where $P(X_1 = k) = 1/6$ for $k \in \{1, 2, 3, 4, 5, 6\}$ .
If $X_1 \in \{4, 5, 6\}$ , no second roll occurs and total sum is $X_1$ :

$X_1 = 4$ : sum is $4 < 6$ .
$X_1 = 5$ : sum is $5 < 6$ .
$X_1 = 6$ : sum is $6 \ge 6$ .
Probability contribution: $P(X_1 = 6) = 1/6 = 6/36$ .

If $X_1 \in \{1, 2, 3\}$ , a second fair roll $X_2$ occurs. We require $X_1 + X_2 \ge 6$ :

If $X_1 = 1$ : $X_2 \ge 5 \implies X_2 \in \{5, 6\}$ (2 outcomes). Probability = $2/36$ .
If $X_1 = 2$ : $X_2 \ge 4 \implies X_2 \in \{4, 5, 6\}$ (3 outcomes). Probability = $3/36$ .
If $X_1 = 3$ : $X_2 \ge 3 \implies X_2 \in \{3, 4, 5, 6\}$ (4 outcomes). Probability = $4/36$ .

Summing all favorable probabilities:

P(\text{Sum} \ge 6) = \frac{6}{36} + \frac{2 + 3 + 4}{36} = \frac{6 + 9}{36} = \frac{15}{36} = \frac{5}{12}

Thus, the probability is $5/12$ .

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2012 Computer Science and Information Technology (CS)
Question: 33 of 65, printed as CS Q.33
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source