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 $\boxed{(B) \frac38}$ .

Video Solution

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

https://www.youtube.com/watch?v=mLrtRuJuYI4 ~David

2000 AMC 8 (Problems • Answer Key • Resources)
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2000 AMC 8 Problems/Problem 21

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