Problem 37 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probabilityor-process-add
A die is rolled 5 times; rolling less than 3 (a 1 or 2) is a 'success'. What is the probability of getting exactly 3 OR exactly 4 successes?
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Answer: 50/243
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Hint 1 of 3
'Exactly 3' and 'exactly 4' can't both happen, so it's an OR process — add their probabilities.
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Hint 2 of 3
You already found \(P(\text{exactly 3}) = 40/243\). Now find \(P(\text{exactly 4})\) the same way.
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Hint 3 of 3
Exactly 4: there are 5 patterns (which roll is the single failure), each with probability \(p^4 q\). So \(P(\text{exactly 4}) = 5 \times (1/3)^4 (2/3)\). Add it to \(40/243\).
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Approach: Binomial probability with an OR (add) over the cases
- 'Exactly 3' and 'exactly 4' successes can't both happen, so add (OR process), with \(p = \frac{1}{3}\) and \(q = \frac{2}{3}\).
- From the previous problem, \(P(\text{exactly }3) = \frac{40}{243}\).
- For exactly 4 successes there are 5 patterns (choose which single roll is the one failure), each worth \(p^4 q\): \(5 \times \left(\frac{1}{3}\right)^4 \times \frac{2}{3} = 5 \times \frac{1}{81} \times \frac{2}{3} = \frac{10}{243}\).
- Add the two cases: \(\frac{40}{243} + \frac{10}{243} = \frac{50}{243}\).
Mark:
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