2000 AMC 8 Problems/Problem 21
Contents
Problem
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
Solution
Divide it into cases:
1) Keiko and Ephriam both get heads: This means that they both roll all tails, so there is only way for this to happen.
2) Keiko and Ephriam both get head: For Keiko, there is only way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are choices for when he can flip the head. So, in total there are ways for this case.
Thus, in total there are ways that work. Since there are choices for each coin flip (Heads or Tails), there are total ways of flipping 3 coins.
Thus, since all possible coin flips of 3 coins are equally likely, the probability is .
~pi_is_3.14
Solution 2
Let be the probability that Keiko gets heads, and let be the probability that Ephriam gets heads.
(Keiko only has one penny!)
(because Ephraim can get HT or TH)
The probability that Keiko gets heads and Ephriam gets heads is . Similarly for head and heads. Thus, we have:
Thus the answer is .
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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