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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== Video Solution == | == Video Solution == | ||
https://youtu.be/a_Tfeb_6dqE Soo, DRMS, NM | 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 10:02, 12 June 2024
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 .
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.