Three unbiased coins are tossed. What is the probability of obtaining exactly two heads (and therefore one tail) among the three outcomes?

Introduction / Context:
When fair coins are tossed, each outcome in the sample space is equally likely. Here, three unbiased coins are tossed, and we seek the probability of getting exactly two heads (and one tail).

Given Data / Assumptions:

• Number of coins = 3.
• Each coin is fair: P(H) = 1/2, P(T) = 1/2.
• All 2^3 outcomes are equiprobable.
• We require exactly two heads.

Concept / Approach:
Use counting with combinations. The number of outcomes with exactly k successes (heads) in n independent Bernoulli trials with p = 1/2 is C(n, k). The probability is C(n, k) / 2^n for fair coins.

Step-by-Step Solution:
Total outcomes = 2^3 = 8.Favorable outcomes for exactly two heads = C(3, 2) = 3 (e.g., HHT, HTH, THH).Probability = 3 / 8.

Verification / Alternative check:
Enumerate all eight outcomes and count the three with two heads to confirm.

Why Other Options Are Wrong:
1/3 and 2/3 confuse counts with probabilities; 3/4 corresponds to “at least two heads” (not correct here); 1/4 corresponds to “all heads” (HHH).

Common Pitfalls:
Forgetting that order matters when counting sequences, but combinations already capture how many distinct placements of heads exist.

Final Answer:
3/8