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I have two coins, one fair, and one unfair which shows heads ¾ of the time....

I have two coins, one fair, and one unfair which shows heads ¾ of the time. I choose a coin at random and flip it 10 times. How many heads do I expect to see? Solve via conditional expectation.

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Answer #1

Expected number of heads, given that the coin is fair, E(X | fair) = 10x0.5 = 5

Expected number of heads, given that the coin is unfair, E(X | unfair) = 10x3/4 = 7.5

Expected number of heads = 0.5x5 + 0.5x0.75

= 6.25

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