Three unbiased coins are tossed simultaneously. What is the probability of getting exactly two heads?

Introduction / Context:
Coin-toss probability questions test counting of outcomes under independence. With unbiased coins, each toss has outcomes H or T with equal likelihood, and combined outcomes are equally likely.

Given Data / Assumptions:

• Three independent, fair coins.
• Sample space size = 2^3 = 8 equiprobable outcomes.
• We want exactly two heads.

Concept / Approach:

• Count favorable outcomes using combinations: choose which 2 of the 3 tosses are heads.
• Probability = favorable / total.

Step-by-Step Solution:

Verification / Alternative check:
List outcomes: HHT, HTH, THH are the only three with exactly two heads; count = 3 out of 8 total outcomes; probability = 3/8.

Why Other Options Are Wrong:

• 1/8 corresponds to exactly three heads (only HHH).
• 2/8 is not the correct count; it simplifies to 1/4.
• 4/8 (= 1/2) counts “at least two heads,” not “exactly two heads.”

Common Pitfalls:

• Confusing “exactly two” with “at least two.”
• Forgetting outcomes are equally likely only if coins are unbiased.

Final Answer:
3/8