Difference between revisions of "2004 AMC 8 Problems/Problem 21"

Revision as of 08:52, 17 February 2023

Problem

Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?

$[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]$

$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$

Solution

An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$ .

$1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{\textbf{(D)}\ \frac23}$

Solution 2

We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is $\boxed{\textbf{(D)}\ \frac23}$

Solution 3

We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we have a $\frac12$ chance of selecting an even number on the first spinner, and the second spinner can be anything, so $\frac12 \cdot 1 = \frac12$ . Next, we consider the probability that we don't get an even on the first spinner but get an even on the second spinner (we can't just consider getting an even for the second spinner and multiply by 1, because we would double count getting even for both). The probability of not getting an even on the first spinner is $\frac12$ and the probability of getting an even on the second spinner is $\frac13$ , so $\frac12 \cdot \frac13 = \frac16$ .

Therefore, the total probability is $\frac12 + \frac16 = \frac46 = \boxed{\textbf{(D)}\ \frac23}$

Video Solution

https://youtu.be/yejMPaQ2uyY Soo, DRMS, NM

@@ Line 30: / Line 30: @@
 ==Solution 3==
-We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we have a \frac12 chance of selecting an even number on the first spinner, and the second spinner can be anything, so \frac12
+We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we have a <cmath>\frac12</cmath> chance of selecting an even number on the first spinner, and the second spinner can be anything, so <cmath>\frac12 \cdot 1 = \frac12</cmath>. Next, we consider the probability that we don't get an even on the first spinner but get an even on the second spinner (we can't just consider getting an even for the second spinner and multiply by 1, because we would double count getting even for both). The probability of not getting an even on the first spinner is <cmath>\frac12</cmath> and the probability of getting an even on the second spinner is <cmath>\frac13</cmath>, so <cmath>\frac12 \cdot \frac13 = \frac16</cmath>.
+Therefore, the total probability is <cmath>\frac12 + \frac16 = \frac46 = \boxed{\textbf{(D)}\ \frac23}</cmath>
 ==Video Solution==

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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