A fair coin is tossed three times and the outcomes are noted. What is the probability of getting exactly two heads?
- (a)2/3
- (b)1/2
- (c)5/8
- (d)3/8
Correct — D, 3/8. Three tosses of a fair coin produce 2³ = 8 equally likely outcomes. Writing them out, exactly two heads occur in HHT, HTH and THH — three of the eight — so the probability is 3/8. The binomial form gives the same figure without listing: ³C₂ × (1/2)² × (1/2) = 3 × 1/8 = 3/8. Note the wording: 'exactly two' excludes HHH, which is why the answer is not 4/8.
- (a)2/3 — 2/3 is not a possible value here at all. Every probability in this experiment has 8 as its denominator, so no answer can be a third.
- (b)1/2 — 1/2 is 4/8, the probability of at least two heads — the three listed outcomes plus HHH. The word 'exactly' removes the fourth.
- (c)5/8 — 5/8 counts too many outcomes. Only three of the eight sequences contain exactly two heads.
Independent repeated trials with two outcomes are binomial. For n tosses of a fair coin the probability of exactly r heads is ⁿCᵣ × (1/2)ⁿ, since every sequence of n tosses has the same probability and ⁿCᵣ of them carry r heads. For n = 3 the counts follow the row 1, 3, 3, 1 of Pascal's triangle, giving probabilities of 1/8, 3/8, 3/8 and 1/8 for zero, one, two and three heads.
With only eight outcomes, listing them is faster and safer than any formula, and it makes the difference between 'exactly two' and 'at least two' visible instead of theoretical. That difference is the whole trap: 1/2 sits in the option list for candidates who include HHH. A last check is that the four probabilities for 0, 1, 2 and 3 heads must add to 1, and 1/8 + 3/8 + 3/8 + 1/8 does.
- Three tosses give 2³ = 8 equally likely outcomes.
- Exactly two heads occurs in HHT, HTH and THH, so the probability is 3/8.
- The binomial formula gives ³C₂ × (1/2)³ = 3/8.
- At least two heads has probability 4/8 = 1/2, which includes HHH.
- The distribution over 0, 1, 2 and 3 heads is 1/8, 3/8, 3/8, 1/8 and sums to 1.
The counts 1, 3, 3, 1 are the third row of Pascal's triangle, and they add to the eight equally likely outcomes.
- Reading 'exactly two' as 'at least two' and answering 1/2.
- Counting HHT and HTH as the same outcome and using 2/8.
- Applying the formula with the wrong combination, ³C₁ instead of ³C₂ — though both happen to give 3 here.
A single-step probability item that repeats almost word for word on CAPF papers, with the 'at least' value planted as a distractor.
A coin is tossed 3 times. The probability of getting exactly 2 heads is
- (a) 1⁄3
- (b) 3⁄8
- (c) 1⁄2
- (d) 5⁄8
Answer(b) 3⁄8
The same question with the same four values, two years earlier and on a paper whose key UPSC published. It is the clearest evidence in this block that CAPF recycles its quantitative items almost unchanged.
- practice — not a real PYQ
A fair coin is tossed four times. What is the probability of getting exactly three heads?
- (a)1/8
- (b)1/4
- (c)3/8
- (d)1/2
Answer(b) 1/4 — ⁴C₃ × (1/2)⁴ = 4/16 = 1/4.
- practice — not a real PYQ
Two dice are thrown together. What is the probability that the sum is 7?
- (a)1/12
- (b)1/9
- (c)1/6
- (d)5/36
Answer(c) 1/6 — six of the 36 outcomes total 7, from 1-6 through to 6-1.