Problem 36 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probabilityor-process-addand-process-multiply
A die is rolled 5 times. Rolling a number less than 3 (a 1 or a 2) counts as a 'success'. What is the probability of getting exactly 3 successes?
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Answer: 40/243
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Hint 1 of 4
First find the chance of a success on ONE roll: 'less than 3' means a 1 or a 2.
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Hint 2 of 4
Any one specific pattern with 3 successes and 2 failures (like S S S F F) has the same probability: multiply 3 success-chances and 2 failure-chances (AND process).
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Hint 3 of 4
Now count how many such patterns exist — that's choosing which 3 of the 5 rolls are the successes. There are 10 of them.
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Approach: Binomial probability — count the patterns, multiply by the per-pattern chance
- A success ('less than 3', a 1 or 2) has probability \(p = \frac{2}{6} = \frac{1}{3}\), and a failure has \(q = \frac{2}{3}\). Any single pattern with 3 successes and 2 failures has probability \(p^3 q^2\) (independent rolls, multiply).
- How many such patterns? That's the number of ways to pick which 3 of the 5 rolls are successes, which is 10. These patterns don't overlap, so add (OR) — i.e. multiply the count by the per-pattern chance.
- So \(10 \times \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^2 = 10 \times \frac{1}{27} \times \frac{4}{9} = \frac{40}{243}\).
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