Difficulty: Easy
Correct Answer: 1/2
Explanation:
Introduction / Context:
This question uses simple probability of coin tosses. It asks for the probability of getting at least two heads when three fair coins are tossed together. The phrase at least two heads means either exactly two heads or exactly three heads, so we sum probabilities of these outcomes.
Given Data / Assumptions:
Concept / Approach:
We can explicitly list all sample space elements for three coin tosses and count the favourable outcomes. Alternatively, we can count outcomes with 0 or 1 head and subtract from 1. Both methods should give the same probability. Here we will list all outcomes directly for clarity.
Step-by-Step Solution:
Total outcomes for three coins = 2^3 = 8.
List them: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Outcomes with at least two heads: HHH, HHT, HTH, THH.
Number of favourable outcomes = 4.
Probability = favourable / total = 4 / 8 = 1 / 2.
Verification / Alternative check:
We can use the complement method. P(0 heads) = P(TTT) = 1/8. Outcomes with exactly one head: HTT, THT, TTH, so P(1 head) = 3/8. Hence P(at least two heads) = 1 - (1/8 + 3/8) = 1 - 4/8 = 4/8 = 1/2. This matches the direct counting method.
Why Other Options Are Wrong:
1/4 would correspond to only 2 favourable outcomes, but there are 4.
3/4 would require 6 favourable outcomes, which is not the case here.
1/3 does not match any natural partition of the 8 outcomes.
Common Pitfalls:
Learners sometimes misinterpret at least two heads as exactly two heads and forget HHH. Others may forget to list all outcomes or miscount them. Systematic listing or using combinations (C(3,2) for exactly two heads and C(3,3) for three heads) helps avoid mistakes.
Final Answer:
Therefore, the probability of getting at least two heads is 1/2.
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