Difference between revisions of "2000 AMC 8 Problems/Problem 21"

Latest revision as of 17:18, 27 October 2024

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

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:

$P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)$

$P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0\cdot\frac{1}{4}$

$P = \frac{3}{8}$

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

Video Solution

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

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

@@ Line 38: / Line 38: @@
 <math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math>
-<math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0</math>
+<math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0\cdot\frac{1}{4}</math>
 <math>P = \frac{3}{8}</math>

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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