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 | + | 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 if 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> | ||
Line 8: | Line 8: | ||
Case 1, 0 heads: | Case 1, 0 heads: | ||
− | The probability of this | + | The probability of this occurring on the coin flip is <math>\frac{1} {4}</math>. The probability that 0 rolls of a die will result in an odd sum is <math>0</math>. |
Case 2, 1 head: | Case 2, 1 head: | ||
− | The probability of this case | + | The probability of this case occurring is <math>\frac{1} {2} \cdot \frac {1} {2} \cdot 2 = \frac {1} {2}.</math> The probability that 1 die results in an odd number is <math>\frac{1} {2}</math>. |
Case 3, 2 heads: | Case 3, 2 heads: | ||
− | The probability of this | + | The probability of this occurring is <math>\frac{1} {4}</math>. The probability that 2 dice result in an odd sum is <math>\frac{1} {2}</math>, because regardless of what we throw on the first die, we have <math>\frac{1} {2}</math> probability that the second die will have the opposite parity. |
Thus, the probability of having an odd sum rolled is <math>\frac{1} {4} \cdot 0 + \frac{1} {2} \cdot \frac{1} {2} + \frac{1} {4} \cdot \frac{1} {2}=\frac{3} {8}\Rightarrow \boxed{A}</math> | Thus, the probability of having an odd sum rolled is <math>\frac{1} {4} \cdot 0 + \frac{1} {2} \cdot \frac{1} {2} + \frac{1} {4} \cdot \frac{1} {2}=\frac{3} {8}\Rightarrow \boxed{A}</math> | ||
+ | |||
+ | == Solution 2 (possibly slightly faster) == | ||
+ | |||
+ | We use complementary counting or subtracting <math>P(\text{Even})</math> from <math>1</math>. We use casework now. | ||
+ | |||
+ | Case <math>1</math>: <math>2</math> Tails. <math>2</math> tails occur with probability <math>\frac{1}{4}</math>, but we will always get an even number, so the overall probability to get an even sum is <math>\frac{1}{4}</math>. | ||
+ | |||
+ | Case <math>2</math>: <math>1</math> Tail: This event occurs with probability <math>\frac{1}{2}</math> and the probability we get an even is <math>\frac{1}{2}</math>, so the overall probability to get an even, in this case, is also <math>\frac{1}{4}</math>. | ||
+ | |||
+ | We know <math>P\text{(Even)}</math> is greater than <math>\frac{1}{2}</math>, so <math>P\text{(Odd)}</math> is less than <math>\frac{1}{2}</math>. | ||
+ | |||
+ | Only <math>\boxed{\text{(A)}\frac{3}{8}}</math> is less than <math>\frac{1}{2}</math>. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2008|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:58, 4 October 2023
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 if 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 occurring 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 occurring is The probability that 1 die results in an odd number is
.
Case 3, 2 heads:
The probability of this occurring is . The probability that 2 dice result in an odd sum is
, because regardless of what we throw on the first die, we have
probability that the second die will have the opposite parity.
Thus, the probability of having an odd sum rolled is
Solution 2 (possibly slightly faster)
We use complementary counting or subtracting from
. We use casework now.
Case :
Tails.
tails occur with probability
, but we will always get an even number, so the overall probability to get an even sum is
.
Case :
Tail: This event occurs with probability
and the probability we get an even is
, so the overall probability to get an even, in this case, is also
.
We know is greater than
, so
is less than
.
Only is less than
.
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 |
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