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A student in Perth conducts a probability simulation to compare two walking routes. Route A has 77 distinct path segments, each equally likely to be chosen.
Route B has 49 distinct path segments, also equally likely. The student uses a digital tool to simulate a single random selection of a path segment from either route, with each route equally likely to be selected first.
What is the exact probability that a randomly selected path segment is from Route A? Express your answer as a simplified fraction.
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Worked solution
- The student selects a route first, with each route equally likely. Therefore, the probability of selecting Route A first is 1/2.
- Given Route A is selected, there are 77 distinct path segments, each equally likely. The probability of selecting any one segment from Route A is 1/77.
- The overall probability of selecting a path segment from Route A is the product of the probability of choosing Route A first and the probability of selecting a segment from Route A given Route A was chosen.
- Multiply the two probabilities: (1/2) × (1/77) = 1/154.
- This result, 1/154, represents the probability of selecting a path segment from Route A in a single random trial.
- The total number of path segments across both routes is 77 (Route A) + 49 (Route B) = 126.
- Since each route is equally likely to be selected first, and segments within a route are equally likely, the probability of selecting any one segment is proportional to the route's selection probability and segment count.
- The total probability mass for Route A is (1/2) × 77 = 77/2.
- The total probability mass for Route B is (1/2) × 49 = 49/2.
- The total probability mass across both routes is 77/2 + 49/2 = 126/2 = 63.
- The probability of selecting a segment from Route A is (77/2) / 63 = 77/(2×63) = 77/126.
- Simplify 77/126 by dividing numerator and denominator by their greatest common divisor, 7: 77 ÷ 7 = 11, 126 ÷ 7 = 18. The simplified fraction is 11/18.
Reference answer: 11/18
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