Difference between revisions of "2004 AMC 10A Problems/Problem 10"
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<math> \mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12 </math> | <math> \mathrm{(A) \ } \frac{29}{128} \qquad \mathrm{(B) \ } \frac{23}{128} \qquad \mathrm{(C) \ } \frac14 \qquad \mathrm{(D) \ } \frac{35}{128} \qquad \mathrm{(E) \ } \frac12 </math> | ||
− | == | + | ==Solution== |
There are <math>4</math> ways that the same number of heads will be obtained; <math>0</math>, <math>1</math>, <math>2</math>, or <math>3</math> heads. | There are <math>4</math> ways that the same number of heads will be obtained; <math>0</math>, <math>1</math>, <math>2</math>, or <math>3</math> heads. | ||
Revision as of 05:09, 14 February 2019
Problem
Coin is flipped three times and coin is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
Solution
There are ways that the same number of heads will be obtained; , , , or heads.
The probability of both getting heads is
The probability of both getting head is
The probability of both getting heads is
The probability of both getting heads is
Therefore, the probabiliy of flipping the same number of heads is:
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.