Difference between revisions of "2015 AMC 10B Problems/Problem 18"
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</math> | </math> | ||
− | ==Solution== | + | ==Solutions== |
− | 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 | + | |
+ | ===Solution 1=== | ||
+ | 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 <math>16</math>, and on the third flip, it is <math>8</math>. Adding these gives <math>\boxed{\mathbf{(D)}\ 56}</math> | ||
+ | ===Solution 2=== | ||
+ | We can simplify the problem first, then move big. Let's say that there are <math>8</math> coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip; | ||
+ | |||
+ | <asy> | ||
+ | filldraw(circle((-5,0),0.35),white); | ||
+ | filldraw(circle((-4,0),0.35),white); | ||
+ | filldraw(circle((-3,0),0.35),white); | ||
+ | filldraw(circle((-2,0),0.35),white); | ||
+ | filldraw(circle((-1,0),0.35),black); | ||
+ | filldraw(circle((-0,0),0.35),black); | ||
+ | filldraw(circle((1,0),0.35),black); | ||
+ | filldraw(circle((2,0),0.35),black); | ||
+ | </asy> | ||
+ | |||
+ | Then, after the second (new heads in blue); | ||
+ | |||
+ | <asy> | ||
+ | filldraw(circle((-5,0),0.35),white); | ||
+ | filldraw(circle((-4,0),0.35),white); | ||
+ | filldraw(circle((-3,0),0.35),blue); | ||
+ | filldraw(circle((-2,0),0.35),blue); | ||
+ | filldraw(circle((-1,0),0.35),black); | ||
+ | filldraw(circle((-0,0),0.35),black); | ||
+ | filldraw(circle((1,0),0.35),black); | ||
+ | filldraw(circle((2,0),0.35),black); | ||
+ | </asy> | ||
+ | |||
+ | And after the third (new head in green); | ||
+ | |||
+ | <asy> | ||
+ | filldraw(circle((-5,0),0.35),white); | ||
+ | filldraw(circle((-4,0),0.35),green); | ||
+ | filldraw(circle((-3,0),0.35),blue); | ||
+ | filldraw(circle((-2,0),0.35),blue); | ||
+ | filldraw(circle((-1,0),0.35),black); | ||
+ | filldraw(circle((-0,0),0.35),black); | ||
+ | filldraw(circle((1,0),0.35),black); | ||
+ | filldraw(circle((2,0),0.35),black); | ||
+ | </asy> | ||
+ | |||
+ | So in total, <math>7</math> of the <math>8</math> coins resulted in heads. Now we have the ratio of <math>\frac{7}{8}</math> of the total coins will end up heads. Therefore, we have <math>\frac{7}{8}\cdot64=\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 19:01, 6 January 2019
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?
Solutions
Solution 1
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 2
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
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.