GATE 2024 DA Q.12: solution and answer

GATE 2024 · DAQ.12 · 1 mark · MCQ

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 $T \cap S$ ?

(A)
0
(B)
0.5
(C)
0.25
(D)
1

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Answer and solution

ANSWER

The probability of $T \cap S$ is $0$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2024 DA (IISc Bengaluru)

Three tosses are made and each toss is a head or a tail, so $(number of heads) + (number of tails) = 3$ .
$T$ means at least 2 heads: $T = {H H H, H H T, H T H, T H H}$ , so $P (T) = \frac{4}{8} = 0.5$ .
$S$ means at least 2 tails: $S = {T T T, T T H, T H T, H T T}$ , so $P (S) = \frac{4}{8} = 0.5$ .
$T \cap S$ needs at least 2 heads and at least 2 tails together, so at least $2 + 2 = 4$ tosses. Only 3 tosses are made, so no outcome can satisfy both. The two outcome sets above also share no element.
Hence $T \cap S = \emptyset$ and $P (T \cap S) = 0.$
Check on independence: $P (T) P (S) = 0.5 \times 0.5 = 0.25 \neq = 0 = P (T \cap S)$ . So $T$ and $S$ are mutually exclusive but not independent. The coins are independent, but the events $T$ and $S$ built from them are not.

KEY IDEA

With 3 tosses, heads plus tails equals 3. Two events that each need at least 2 of the 3 tosses cannot both occur, so they are mutually exclusive and the intersection is empty.

WATCH OUT FOR

Multiplying $P (T) P (S) = 0.25$ as if $T$ and $S$ were independent, and picking (C). Independent coins do not make $T$ and $S$ independent. In fact $S$ is determined by the same tosses and cannot occur when $T$ does.

From the GATE 2024 DA paper. Every question of this paper · Download the full paper · All 5 Counting & axioms questions, with notes

GATE 2024 DA Q.12: solution and answer

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