Difference between revisions of "2015 AMC 10B Problems/Problem 18"
(→Solution) |
Firebolt360 (talk | contribs) (→Solution) |
||
Line 9: | Line 9: | ||
</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 20:01, 6 January 2019
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.