Counting and Probability

GATE DA previous year questions · 2 questions

GATE DA 2025 | Question 45

MCQ · 2 marks · −0.66

A random experiment consists of throwing 100 fair dice, each die having six faces numbered 1 to 6. An event \(A\) represents the set of all outcomes where at least one of the dice shows a 1. Then, \(P(A) =\)

0 1 \(1 - \left(\frac{5}{6}\right)^{100}\) \(\left(\frac{5}{6}\right)^{100}\)

Explanation

Answer: (C)

Use the complement: the event fails only when no die shows a 1.

Each die avoids 1 with probability \(\frac{5}{6}\), and the 100 throws are independent, so \(P(\text{no 1 at all}) = \left(\frac{5}{6}\right)^{100}\). Therefore

\[P(A) = 1 - \left(\frac{5}{6}\right)^{100}.\]

Option D is the probability of the complement. Options A and B fail because both \(A\) and its complement can happen, so \(P(A)\) lies strictly between 0 and 1, though very close to 1.

Answer: C

GATE DA 2024 | Question 12

MCQ · 1 mark · −0.33

Three fair coins are tossed independently. \(T\) is the event that two or more tosses result in heads. \(S\) is the event that two or more tosses result in tails.

What is the probability of the event \(T \cap S\) ?

0 0.5 0.25 1

Explanation

Answer: (A)

\(T \cap S\) needs at least two heads and at least two tails, which takes at least four tosses. With three coins, two or more heads leave at most one tail, so the two events can never happen together.

\(T\) and \(S\) are mutually exclusive (in fact complementary, since every outcome of three tosses has a majority of heads or a majority of tails), so \(P(T \cap S) = 0\). The value 0.5 is \(P(T)\) alone.

Answer: A