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

(Created page with "Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is <math> \text{(A)}\ \frac{1}{4}\qquad\text{(B...")
 
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==Problem==
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Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
 
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
  
 
<math> \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} </math>
 
<math> \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} </math>
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==Solution==
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Divide it into <math>2</math> cases:
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1) Keiko and Ephriam both get <math>0</math> heads:
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This means that they both roll all tails, so there is only <math>1</math> way for this to happen.
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2) Keiko and Ephriam both get <math>1</math> head:
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For Keiko, there is only <math>1</math> way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are <math>2</math> choices for when he can flip the head. So, in total there are <math>2 \cdot 1 = 2</math> ways for this case.
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Thus, in total there are <math>3</math> ways that work. Since there are <math>2</math> choices for each coin flip (Heads or Tails), there are <math>2^3 = 8</math> total ways of flipping 3 coins.
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Thus, since all possible coin flips of 3 coins are equally likely, the probability is <math>\boxed{(B) \frac38}</math>.
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~pi_is_3.14
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==Solution 2==
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Let <math>K(n)</math> be the probability that Keiko gets <math>n</math> heads, and let <math>E(n)</math> be the probability that Ephriam gets <math>n</math> heads.
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<math>K(0) = \frac{1}{2}</math>
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<math>K(1) = \frac{1}{2}</math>
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<math>K(2) = 0</math>  (Keiko only has one penny!)
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<math>E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
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<math>E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}</math>  (because Ephraim can get HT or TH)
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<math>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
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The probability that Keiko gets <math>0</math> heads and Ephriam gets <math>0</math> heads is <math>K(0)\cdot E(0)</math>.  Similarly for <math>1</math> head and <math>2</math> heads.  Thus, we have:
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<math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math>
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<math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0</math>
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<math>P = \frac{3}{8}</math>
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Thus the answer is <math>\boxed{B}</math>.
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== Video Solution ==
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https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM
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==See Also==
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{{AMC8 box|year=2000|num-b=20|num-a=22}}
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{{MAA Notice}}

Revision as of 17:21, 8 October 2022

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

See Also

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

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