2015 AMC 10B Problems/Problem 18
Contents
Problem
Johann has 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?
Solution 1
We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
Then, after the second (new heads in blue);
And after the third (new head in green);
So in total, of the coins resulted in heads. Now we have the ratio of of the total coins will end up heads. Therefore, we have
Solution 2 (Efficient)
Every time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is , on the second flip is , and on the third flip, it is . Adding these gives
Solution 3
Every time the coins are flipped, each of them has a probability of being tails. Doing this times, of them will be tails. .
~Lcz
Solution 4
(Similar to solution 2)
Notice how:
The expected number of heads for the first flip is simply , since each coin has a 1 in 2 chance of being heads. Then, we are left with coins. Then, half of these coins will be heads again, which leaves us with coins. Then, half of these coins will be heads again, which leaves us with coins.
Hence, the expected number of heads is simply:
~yk2007
Video Solution
~savannahsolver
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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