Problem 31 · AMC 8 Stretch
Core
Counting & Probability
and-process-multiply
Helen tosses a coin 6 times. Find the probability that she gets heads on the first 3 tosses and tails on the last 3 tosses (in that exact order: H H H T T T).
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Answer: 1/64
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Hint 1 of 3
This asks for one exact sequence H H H T T T, not 'three of each in any order'. Each toss is independent.
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Hint 2 of 3
Multiply the chance of the required result on every one of the 6 tosses.
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Hint 3 of 3
Each toss has probability \(1/2\), so it's \((1/2)\) multiplied 6 times.
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Approach: AND process for one specific sequence
- An exact order H H H T T T is required. Each toss is independent with probability \(\frac{1}{2}\), so multiply (AND process).
- So \(\left(\frac{1}{2}\right)^6 = \frac{1}{64}\). Because we want one specific order, there is no 'choosing' step — just the single product.
Mark:
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