Difference between revisions of "2008 AMC 10B Problems/Problem 16"
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We use complementary counting or subtracting <math>P(\text{Even})</math> from <math>1</math>. We use casework now. | 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 is <math>\frac{1}{4} | + | 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 is <math>\frac{1}{4}</math>. |
− | Case < | + | Case <math>2</math>: <math>1</math> Tail: <math>This event occurs with probability </math>\frac{1}{2}<math> and the probability we get on 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 the <math>P\text{(Even)}< | + | We know the </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}< | + | Only </math>\boxed\text{(A)}\frac{3}{8}<math> is less than </math>\frac{1}{2}$ |
==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}} | {{MAA Notice}} |
Revision as of 12:54, 25 December 2019
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 one die results as 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 (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 is
.
Case :
Tail:
\frac{1}{2}
\frac{1}{2}
\frac{1}{4}
P\text{(Even)}
\frac{1}{2}
P\text{(Odd)}
\frac{1}{2}$.
Only$ (Error compiling LaTeX. Unknown error_msg)\boxed\text{(A)}\frac{3}{8}\frac{1}{2}$
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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