Mathematics — Standard (041) — CBSE Class 10 board question

Sample Space: When 3 coins are tossed, total possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8

(i) At least one head:
Favourable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH = 7

$$P(\text{at least one head}) = \frac{7}{8}$$

(ii) Exactly one tail:
Favourable outcomes: HHT, HTH, THH = 3

$$P(\text{exactly one tail}) = \frac{3}{8}$$

(iii) Two heads and one tail:
Favourable outcomes: HHT, HTH, THH = 3

$$P(\text{two heads and one tail}) = \frac{3}{8}$$

Source: Chapter 14, Probability — A Theoretical Approach

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• Always list the full sample space first — examiners expect to see it.
• Note that (ii) and (iii) describe the same event (2H + 1T = exactly 1T), so both answers are $\frac{3}{8}$.
• Use the formula: $P(E) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}$.
• "At least one head" is easier via complement: $1 - P(\text{no head}) = 1 - \frac{1}{8} = \frac{7}{8}$ — either method is acceptable.