Difference between revisions of "2015 AMC 10B Problems/Problem 18"
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+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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Revision as of 09:13, 4 August 2020
Contents
[hide]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 move big. 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
Video Solution
~savannahsolver
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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