Expected number of free throws made

Question

A player takes 5 free throws. Because of fatigue, the make probabilities are 0.9, 0.8, 0.7, 0.6, and 0.5 in order. On average, how many of the 5 free throws does the player make?

Accepted Answer

Put one indicator on each shot: Xi = 1 if shot i goes in. Then E[Xi] equals that shot's make probability. The number of makes is X = X1 + \cdots + X5, and by linearity of expectation E[X] = 0.9 + 0.8 + 0.7 + 0.6 + 0.5 = 3.5. The shots did not need equal probabilities; linearity simply adds whatever the individual chances are. Takeaway. Expected makes equal the sum of the per-shot make probabilities: 0.9 + 0.8 + 0.7 + 0.6 + 0.5 = 3.5.

Expected number of free throws made

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