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

Latest revision as of 12:54, 27 August 2021

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

Video Solution

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

~savannahsolver

@@ Line 11: / Line 11: @@
 ==Solution 1==
-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;
+We can simplify the problem first, then apply reasoning to the original problem. 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>
@@ Line 59: / Line 59: @@
 ~Lcz
+==Solution 4==
+(Similar to solution 2)
+Notice how:
+<math>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})</math>
+The expected number of heads for the first flip is simply <math>64 \cdot \frac{1}{2}</math>, since each coin has a 1 in 2 chance of being heads. Then, we are left with <math>64 - 32 = 32</math> coins. Then, half of these coins will be heads again, which leaves us with <math>32 - 16 = 16</math> coins. Then, half of these coins will be heads again, which leaves us with <math>16 - 8 = 8</math> coins.
+Hence, the expected number of heads is simply:
+<math>32 + 16 + 8 = \boxed{56} \implies D</math>
+~yk2007
 ==Video Solution==

2015 AMC 10B (Problems • Answer Key • Resources)
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