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> | 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> | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2015|ab=B|num-b=17|num-a=19}} | ||
| + | {{MAA Notice} | ||
Revision as of 18:05, 4 March 2015
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
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 | ||
{{MAA Notice}
