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Difference between revisions of "2015 AMC 10B Problems/Problem 18"

(Created page with "==Problem== Johann has <math>64</math> 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...")
 
(12 intermediate revisions by 6 users not shown)
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</math>
 
</math>
  
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==Solution==
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==Solution 1==
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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>
+
 
 +
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>
 +
 
 +
==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 <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 3==
 +
Every time the coins are flipped, each of them has a <math>1/2</math> probability of being tails. Doing this <math>3</math> times, <math>1/8</math> of them will be tails. <math>64-64*1/8=</math><math>\boxed{\mathbf{(D)}\ 56}</math>.
 +
 
 +
~Lcz
 +
 
 +
==Video Solution==
 +
https://youtu.be/0uCMSH7-Ubk
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=B|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 10:13, 4 August 2020

Problem

Johann has $64$ 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?

$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution 1

We can simplify the problem first, then move big. Let's say that there are $8$ 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, $7$ of the $8$ coins resulted in heads. Now we have the ratio of $\frac{7}{8}$ of the total coins will end up heads. Therefore, we have $\frac{7}{8}\cdot64=\boxed{\mathbf{(D)}\ 56}$

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 $32$, on the second flip is $16$, and on the third flip, it is $8$. Adding these gives $\boxed{\mathbf{(D)}\ 56}$

Solution 3

Every time the coins are flipped, each of them has a $1/2$ probability of being tails. Doing this $3$ times, $1/8$ of them will be tails. $64-64*1/8=$$\boxed{\mathbf{(D)}\ 56}$.

~Lcz

Video Solution

https://youtu.be/0uCMSH7-Ubk

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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