# 2015 AMC 10B Problems/Problem 18

## 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 apply reasoning to the original problem. 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

## Solution 4

(Similar to solution 2)

Notice how:

$E(\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\text{Heads on 1st flip}) + E(\text{Heads on 2nd flip}) + E(\text{Heads on 3rd flip})$

The expected number of heads for the first flip is simply $64 \cdot \frac{1}{2}$, since each coin has a 1 in 2 chance of being heads. Then, we are left with $64 - 32 = 32$ coins. Then, half of these coins will be heads again, which leaves us with $32 - 16 = 16$ coins. Then, half of these coins will be heads again, which leaves us with $16 - 8 = 8$ coins.

Hence, the expected number of heads is simply:

$32 + 16 + 8 = \boxed{56} \implies D$

~yk2007

~savannahsolver

## See Also

 2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 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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