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

(→‎Solution)
(→‎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 is is <math>16</math>, and on the third flip it is <math>8</math>. Adding these gives <math>\boxed{\mathbf{(D)}\ 56}</math>
+
 
 +
===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

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?

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

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

Solution 2

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 $\frac{7}{8}\cdot64=\boxed{\mathbf{(D)}\ 56}$

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

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