# 2000 AMC 8 Problems/Problem 21

## Problem

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is $\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

## Solution

Divide it into $2$ cases:

1) Keiko and Ephriam both get $0$ heads: This means that they both roll all tails, so there is only $1$ way for this to happen.

2) Keiko and Ephriam both get $1$ head: For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.

Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.

Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{(B) \frac38}$.

~pi_is_3.14

## Solution 2

Let $K(n)$ be the probability that Keiko gets $n$ heads, and let $E(n)$ be the probability that Ephriam gets $n$ heads. $K(0) = \frac{1}{2}$ $K(1) = \frac{1}{2}$ $K(2) = 0$ (Keiko only has one penny!) $E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$ $E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}$ (because Ephraim can get HT or TH) $E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$

The probability that Keiko gets $0$ heads and Ephriam gets $0$ heads is $K(0)\cdot E(0)$. Similarly for $1$ head and $2$ heads. Thus, we have: $P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)$ $P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0$ $P = \frac{3}{8}$

Thus the answer is $\boxed{B}$.

## Video Solution

https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM

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