If a fair coin is tossed 3 times, what is the probability of getting at least two heads?

Analyze the question: We toss a fair coin 3 times and want P(at least two heads), meaning exactly 2 heads or exactly 3 heads.

Concept: For a fair coin tossed n times, there are 2ⁿ equally likely outcomes. The number of ways to get r heads is ⁿCᵣ.

Solution:

Total outcomes = 2³ = 8

The outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

• Exactly 2 heads: ³C₂ = 3 (HHT, HTH, THH)
• Exactly 3 heads: ³C₃ = 1 (HHH)

Favourable outcomes = 3 + 1 = 4

P(at least 2 heads) = 4/8 = 1/2

Check with the binomial formula: P(2 heads) = 3 × (1/2)³ = 3/8 P(3 heads) = 1 × (1/2)³ = 1/8 Total = 3/8 + 1/8 = 4/8 = 1/2 ✔

Answer: The probability is 1/2, which matches option D.

Tip: For 3 tosses of a fair coin, the outcomes are symmetric. "At least 2 heads" and "at most 1 head" (that is, at least 2 tails) have equal probability, so each must be exactly 1/2. Whenever the number of tosses is odd and you want "more than half" heads, the answer is 1/2.