Difference between revisions of "2008 AMC 10B Problems/Problem 16"
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==Problem== | ==Problem== | ||
− | Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd?(Note that is no die is rolled, the sum is 0.) | + | Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that is no die is rolled, the sum is 0.) |
<math>\mathrm{(A)}\ {{{\frac{3} {8}}}} \qquad \mathrm{(B)}\ {{{\frac{1} {2}}}} \qquad \mathrm{(C)}\ {{{\frac{43} {72}}}} \qquad \mathrm{(D)}\ {{{\frac{5} {8}}}} \qquad \mathrm{(E)}\ {{{\frac{2} {3}}}}</math> | <math>\mathrm{(A)}\ {{{\frac{3} {8}}}} \qquad \mathrm{(B)}\ {{{\frac{1} {2}}}} \qquad \mathrm{(C)}\ {{{\frac{43} {72}}}} \qquad \mathrm{(D)}\ {{{\frac{5} {8}}}} \qquad \mathrm{(E)}\ {{{\frac{2} {3}}}}</math> |
Revision as of 15:59, 25 January 2009
Problem
Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that is no die is rolled, the sum is 0.)
Solution
We consider 3 cases based on the outcome of the coin:
Case 1, 0 heads:
The probability of this occuring on the coin flip is . The probability that 0 rolls of a die will result in an odd sum is
.
Case 2, 1 head:
The probability of this case occuring is . The proability that one die results as an odd number is
.
Case 3, 2 heads:
The probability of this occuring is . The probability that 2 die result in an odd sum is
Thus, the probability of having an odd sum rolled is
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |