Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 10B Problems/Problem 16"

(→‎Solution)
(→‎Solution)
 
(9 intermediate revisions by 2 users not shown)
Line 11: Line 11:
  
 
Case 2, 1 head:
 
Case 2, 1 head:
−
The probability of this case occurring is <math>\frac{1} {2}</math>. The probability that one die results as an odd number is <math>\frac{1} {2}</math>.
+
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:
Line 18: Line 18:
 
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 (slightly faster) ==
+
== Solution 2 (possibly slightly faster) ==
  
 
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 sum is <math>\frac{1}{4}</math>.
  
−
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>
+
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 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>.
+
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>
+
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}}
 
{{MAA Notice}}

Latest revision as of 21:58, 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.)

$\mathrm{(A)}\ {{{\frac{3} {8}}}} \qquad \mathrm{(B)}\ {{{\frac{1} {2}}}} \qquad \mathrm{(C)}\ {{{\frac{43} {72}}}} \qquad \mathrm{(D)}\ {{{\frac{5} {8}}}} \qquad \mathrm{(E)}\ {{{\frac{2} {3}}}}$

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 $\frac{1} {4}$. The probability that 0 rolls of a die will result in an odd sum is $0$.

Case 2, 1 head: The probability of this case occurring is $\frac{1} {2} \cdot \frac {1} {2} \cdot 2 = \frac {1} {2}.$ The probability that 1 die results in an odd number is $\frac{1} {2}$.

Case 3, 2 heads: The probability of this occurring is $\frac{1} {4}$. The probability that 2 dice result in an odd sum is $\frac{1} {2}$, because regardless of what we throw on the first die, we have $\frac{1} {2}$ probability that the second die will have the opposite parity.

Thus, the probability of having an odd sum rolled is $\frac{1} {4} \cdot 0 + \frac{1} {2} \cdot \frac{1} {2} + \frac{1} {4} \cdot \frac{1} {2}=\frac{3} {8}\Rightarrow \boxed{A}$

Solution 2 (possibly slightly faster)

We use complementary counting or subtracting $P(\text{Even})$ from $1$. We use casework now.

Case $1$: $2$ Tails. $2$ tails occur with probability $\frac{1}{4}$, but we will always get an even number, so the overall probability to get an even sum is $\frac{1}{4}$.

Case $2$: $1$ Tail: This event occurs with probability $\frac{1}{2}$ and the probability we get an even is $\frac{1}{2}$, so the overall probability to get an even, in this case, is also $\frac{1}{4}$.

We know $P\text{(Even)}$ is greater than $\frac{1}{2}$, so $P\text{(Odd)}$ is less than $\frac{1}{2}$.

Only $\boxed{\text{(A)}\frac{3}{8}}$ is less than $\frac{1}{2}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_16&oldid=199142"