Difference between revisions of "2000 AMC 8 Problems/Problem 21"
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Thus, since all possible coin flips of 3 coins are equally likely, the probability is <math>\boxed{(B) \frac38}</math>. | 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== | ==Solution 2== |
Revision as of 15:06, 24 November 2023
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 .
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.