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ISEE Upper Level Mathematics Achievement

ISEE Upper Level Mathematics Achievement Question of the Day

Practice ISEE Upper Level Mathematics Achievement with the production-style question-of-the-day selection for this public URL.

Question 1

A coin is flipped three times. What is the probability of getting exactly two heads, given that at least one head occurred?

  1. 14\frac{1}{4}41​
  2. 37\frac{3}{7}73​
  3. 38\frac{3}{8}83​
  4. 12\frac{1}{2}21​
Explanation: This is a conditional probability problem, which asks for the probability of one event happening given that another event has already occurred. When you see "given that" in a probability question, you need to use the conditional probability formula or adjust your sample space. First, let's find all possible outcomes when flipping a coin three times. There are 23=82^3 = 823=8 total outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Since we're given that at least one head occurred, we can eliminate TTT from our sample space. This leaves us with 7 favorable outcomes for the condition: HHH, HHT, HTH, HTT, THH, THT, TTH. Now we need exactly two heads from these 7 remaining outcomes. Looking at our list: HHT, HTH, and THH each contain exactly two heads. That's 3 outcomes with exactly two heads. Therefore, the probability is 37\frac{3}{7}73​. Choice A (14\frac{1}{4}41​) incorrectly uses the original sample space of 8 outcomes instead of the reduced space of 7. Choice C (38\frac{3}{8}83​) correctly identifies 3 favorable outcomes but fails to account for the conditional restriction, using all 8 original outcomes as the denominator. Choice D (12\frac{1}{2}21​) likely comes from incorrectly thinking there are only 6 remaining outcomes or miscounting the favorable cases. Remember: in conditional probability problems, always adjust your sample space to include only the outcomes that satisfy the given condition, then find your target event within that restricted space.