🇺🇸 AMC 8 ⇄ switch contest
1996 AMC 8 Stretch

Problem 40

Problem 40 · AMC 8 Stretch Stretch
Counting & Probability binomial-probability
Find the probability of getting exactly 4 heads in 6 tosses of a fair coin.
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Answer: 15/64
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Hint 1 of 3
Each toss is heads with probability \(1/2\). There are \(2^6 = 64\) equally likely head/tail sequences in all.
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Hint 2 of 3
Count how many of those sequences have exactly 4 heads — that's choosing which 4 of the 6 tosses are heads.
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Hint 3 of 3
There are 15 ways to choose which 4 tosses are heads. Divide by 64.
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Approach: Favorable sequences over all sequences
  1. All 6 tosses give \(2^6 = 64\) equally likely sequences.
  2. The number with exactly 4 heads is the number of ways to choose which 4 of the 6 tosses are heads, which is 15.
  3. So the probability is \(\frac{15}{64}\).
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