GATE 2023 EE GA Q.3: solution and answer

GATE 2023 · EEGA Q.3 · 1 mark · MCQ

Given a fair six-faced dice where the faces are labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, what is the probability of getting a ‘1’ on the first roll of the dice and a ‘4’ on the second roll?

(A)
$\frac{1}{36}$
(B)
$\frac{1}{6}$
(C)
$\frac{5}{6}$
(D)
$\frac{1}{3}$

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Answer and solution

ANSWER

The probability of a '1' on the first roll and a '4' on the second roll is $\frac{1}{36}$ , option (A).

Official answer: ASource: Official answer key, GATE 2023 EE (IIT Kanpur)

A fair die gives each face probability $\frac{1}{6}$ on any single roll, so $P (1 on first roll) = \frac{1}{6}$ and $P (4 on second roll) = \frac{1}{6}$ .
Separate rolls are independent: the result of the first roll does not change the probabilities for the second. The probability that both events happen is therefore the product $P (A \cap B) = P (A) P (B)$ .
$P = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
Check by counting: two rolls give $6 \times 6 = 36$ equally likely ordered outcomes, and exactly one of them, $(1, 4)$ , is favourable. So $P = \frac{1}{36}$ , which is option (A).

KEY IDEA

When events are independent, the probability that all of them happen ("and") is the product of their separate probabilities. For two rolls of a fair die the sample space is $6 \times 6 = 36$ equally likely ordered pairs, and a fixed ordered result such as $(1, 4)$ is exactly one of them.

WATCH OUT FOR

Adding the two probabilities instead of multiplying them ( $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ , option D) confuses "and" with "or". Picking $\frac{1}{6}$ (option B) looks at only one roll. Another slip is counting $(4, 1)$ as favourable as well and getting $\frac{2}{36}$ . The order is fixed here: '1' first, then '4'.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 7 Statistics & probability questions, with notes

GATE 2023 EE GA Q.3: solution and answer

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