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Difference between revisions of "2000 AMC 8 Problems/Problem 21"
 
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==Solution==
 
==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.
 +
 +
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.
 +
 +
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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==Solution 2==
  
 
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.
 
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>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
 
<math>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math>
  
−
The probability that Keiko gets <math>0</math> heads and Ephriam gets <math>0</math> heads is <math>K(0)\cdot E(0)</math>.  Simiarly for <math>1</math> head and <math>2</math> heads.  Thus, we have:
+
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:
  
 
<math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math>
 
<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>
 
<math>P = \frac{3}{8}</math>
  
 
Thus the answer is <math>\boxed{B}</math>.
 
Thus the answer is <math>\boxed{B}</math>.
 +
 +
==Solution 3==
 +
Keiko can only get 1 head or one tail. That means Ephriam has to also get 1 head or one tail. There are 3 cases <math>HHT, HTH, TTT</math> and each have a <math>\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}</math> chance. Adding all 3 ways makes a total of <math>\boxed{(B)\frac38}</math>
 +
 +
~RandomMathGuy500
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 +
 +
== Video Solution ==
 +
https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM
 +
 +
https://www.youtube.com/watch?v=mLrtRuJuYI4    ~David
  
 
==See Also==
 
==See Also==

Latest revision as of 02:00, 29 December 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 $\boxed{B}$.

Solution 3

Keiko can only get 1 head or one tail. That means Ephriam has to also get 1 head or one tail. There are 3 cases $HHT, HTH, TTT$ and each have a $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$ chance. Adding all 3 ways makes a total of $\boxed{(B)\frac38}$

~RandomMathGuy500


Video Solution

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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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