Problem 12 · AMC 8 Stretch
Stretch
Counting & Probability
visual-representationaccount-for-all-possibilities
Flip a fair coin \(4\) times. The chances of getting exactly \(0, 1, 2, 3, 4\) heads come from the counts \(1, 4, 6, 4, 1\) out of \(16\). What is the most likely number of heads?
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Answer: 2 heads (probability 6/16 = 3/8)
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Hint 1 of 4
Every sequence of H's and T's in \(4\) flips is equally likely. How many sequences are there in total? (\(2 \times 2 \times 2 \times 2\).)
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Hint 2 of 4
The chance of exactly \(k\) heads is (number of sequences with \(k\) heads) divided by the total. So you need to count sequences for each \(k\).
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Hint 3 of 4
Use the numbers from Pascal's triangle row for \(4\): \(1, 4, 6, 4, 1\). (There are \(6\) ways to place \(2\) heads among \(4\) flips, and so on.)
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Approach: Count sequences with Pascal's triangle; the distribution peaks in the middle
- Each set of \(4\) flips is one of \(2^4 = 16\) equally likely sequences, so \(P(k \text{ heads}) = \frac{\text{sequences with } k \text{ heads}}{16}\).
- The counts come from Pascal's triangle (row 4): \(1, 4, 6, 4, 1\), giving probabilities \(\frac{1}{16}, \frac{4}{16}, \frac{6}{16}, \frac{4}{16}, \frac{1}{16}\).
- Check the total: \(1 + 4 + 6 + 4 + 1 = 16\), so the probabilities add to \(\frac{16}{16} = 1\).
- The largest count is \(6\), at \(k = 2\). So the most likely outcome is exactly \(2\) heads, with probability \(\frac{6}{16} = \frac{3}{8}\).
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