Year 7 Chance experiments and simulations — Medium exercise 8

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In a simulation of a walking route in Canberra, two paths are compared: Path A has a 76/97 probability of being chosen at each decision point, and Path B has the complementary probability. The simulation runs 9700 trials, with each trial representing a single walk from start to finish.

What is the exact expected number of times Path A will be chosen? Express your answer as an exact fraction.

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Choose 0 or 1 with equal probability: 0 means tails and 1 means heads. Repeat independent trials and count outcomes. No experiment results are asserted here.

Outcomes 5 and 6 are greater than 4: 2 favourable outcomes out of 6, so probability = 1/3.

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Worked solution

The probability of choosing Path A at each decision point is given as 76/97.
The simulation runs 9700 independent trials, each representing one complete walk.
The expected number of times Path A is chosen is calculated by multiplying the probability of choosing Path A by the total number of trials.
Substitute the values: expected count = (76/97) × 9700.
Simplify the expression: (76/97) × 9700 = 76 × (9700 / 97).
Divide 9700 by 97: 9700 ÷ 97 = 100, since 97 × 100 = 9700.
Now compute: 76 × 100 = 7600.
The result is an integer, so the exact expected number is 7600.
Verify: 76/97 × 9700 = 76 × 100 = 7600. The fraction simplifies cleanly because 9700 is a multiple of 97.
The expected number of times Path A is chosen is exactly 7600.

Reference answer: 7600

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