2015 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2015 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10B solutions, or check the answer key.

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Concepts:expected valuecomplementary probability

Difficulty rating: 1070

18.

Johann has 6464 fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?

32 32

40 40

48 48

56 56

64 64

Solution:

A coin ends as tails if and only if it has 33 flips that are tails, which happens with probability 18.\frac 18. Thus, the probability of any coin being heads is 78.\frac 78 .

As the probability that a given coin flip is 78,\frac 78, and there are 6464 coin flips in total, the expected number of coins that are now heads is: 6478=56.64 \cdot \dfrac 78 = 56.

Thus, the correct answer is D .

Problem 18 in Other Years