Difference between revisions of "2015 AMC 10B Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | The expected number of heads on the first flip is <math>32</math>, on the second flip is is <math>16</math>, and on the third flip it is <math>8</math>. Adding these gives <math>\boxed{\mathbf{(D)}\ 56}</math> | + | 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 <math>32</math>, on the second flip is is <math>16</math>, and on the third flip it is <math>8</math>. Adding these gives <math>\boxed{\mathbf{(D)}\ 56}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2015|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:59, 1 January 2016
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
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 is 
, and on the third flip it is 
. Adding these gives 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
