Difference between revisions of "2015 AMC 8 Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? | + | Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? |
<math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}</math> | <math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}</math> | ||
==Solutions== | ==Solutions== | ||
+ | ===Solution 1.1=== | ||
+ | (This solution is similar to Solution 2.) | ||
+ | Let's make this a problem with boxes. | ||
+ | |||
+ | In total, there are 9 products derived from these numbers (because 3 numbers per box). This would be our denominator. | ||
+ | |||
+ | Every time a spinner lands on 2, we get an even product. These are <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(2,3)</math>, and finally <math>(3,2)</math>. | ||
+ | |||
+ | Going back, we see there are <math>3\cdot3=9</math> possible combinations, and we have <math>5</math> evens, the final answer is | ||
+ | |||
+ | <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
+ | |||
===Solution 1=== | ===Solution 1=== | ||
− | |||
− | |||
− | |||
You can also make this problem into a spinner problem. You have the first spinner with <math>3</math> equally divided | You can also make this problem into a spinner problem. You have the first spinner with <math>3</math> equally divided | ||
− | sections | + | sections: <math>1, 2</math>, and <math>3</math>. You make a second spinner that is identical to the first, with <math>3</math> equal sections of |
− | <math>1</math>,<math>2</math>, and <math>3</math>. If the first spinner lands on <math>1</math> | + | <math>1</math>,<math>2</math>, and <math>3</math>. If the first spinner lands on <math>1</math>, it must land on two for the result to be even. You write down the first |
− | combination of numbers <math>(1,2)</math>. Next, if the spinner lands on <math>2</math>, it can land on any number on the second | + | combination of numbers: <math>(1,2)</math>. Next, if the spinner lands on <math>2</math>, it can land on any number on the second |
− | spinner. We now have the combinations of <math>(1,2) ,(2,1), (2,2), (2,3)</math>. Finally, if the first spinner ends on <math>3</math>, we | + | spinner. We now have the combinations of <math>(1,2) ,(2,1), (2,2),</math> and <math>(2,3)</math>. Finally, if the first spinner ends on <math>3</math>, we |
− | have <math>(3,2).</math> Since there are <math>3 | + | have <math>(3,2).</math> Since there are <math>3\cdot3=9</math> possible combinations, and we have <math>5</math> evens, the final answer is |
<math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
===Solution 3=== | ===Solution 3=== | ||
− | We can also list out the numbers. Box A has chips <math>1</math>, <math>2</math>, and <math>3</math>, and Box B also has chips <math>1</math>, <math>2</math>, and <math>3</math>. Chip <math>1</math>(from Box A) | + | We can also list out the numbers. Box A has chips <math>1</math>, <math>2</math>, and <math>3</math>, and Box B also has chips <math>1</math>, <math>2</math>, and <math>3</math>. Chip <math>1</math> (from Box A) |
could be with 3 partners from Box B. This is also the same for chips <math>2</math> and <math>3</math> from Box A. <math>3+3+3=9</math> total sums. Chip <math>1</math> could be | could be with 3 partners from Box B. This is also the same for chips <math>2</math> and <math>3</math> from Box A. <math>3+3+3=9</math> total sums. Chip <math>1</math> could be | ||
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There are <math>3</math> chips we can choose from in the 1st box, and <math>3</math> chips we can choose from in the 2nd box. We do <math>3*3</math>, and get <math>9</math>. | There are <math>3</math> chips we can choose from in the 1st box, and <math>3</math> chips we can choose from in the 2nd box. We do <math>3*3</math>, and get <math>9</math>. | ||
Now - to find the numerator: Desired choices. | Now - to find the numerator: Desired choices. | ||
− | To get an even number, we need to pick 2 from at least one of the boxes. There are <math>2</math> choices as to finding which box we will draw the 2 from. Then we have <math>3</math> choices from the other box to pick any of the other chips, <math>1, 2, 3</math>. | + | To get an even number, we need to pick 2 from at least one of the boxes. There are <math>2</math> choices as to finding which box we will draw the 2 from. Then, we have <math>3</math> choices from the other box to pick any of the other chips, <math>1, 2,</math> and <math>3</math>. |
<math>\frac{3 \cdot2}{9} = \frac{6}{9}</math> | <math>\frac{3 \cdot2}{9} = \frac{6}{9}</math> | ||
− | However, we are over counting | + | |
+ | However, we are over counting the <math>(2,2)</math> configuration twice, and so, we subtract that one configuration from our total. | ||
<math>\frac{6}{9} - \frac{1}{9}</math>. | <math>\frac{6}{9} - \frac{1}{9}</math>. | ||
Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
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~MiracleMaths | ~MiracleMaths | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/oODaaTOHemg | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/PBBrT9iVCVU | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 16:54, 6 January 2024
Contents
Problem
Each of two boxes contains three chips numbered ,
,
. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?
Solutions
Solution 1.1
(This solution is similar to Solution 2.) Let's make this a problem with boxes.
In total, there are 9 products derived from these numbers (because 3 numbers per box). This would be our denominator.
Every time a spinner lands on 2, we get an even product. These are ,
,
,
, and finally
.
Going back, we see there are possible combinations, and we have
evens, the final answer is
.
Solution 1
You can also make this problem into a spinner problem. You have the first spinner with equally divided
sections: , and
. You make a second spinner that is identical to the first, with
equal sections of
,
, and
. If the first spinner lands on
, it must land on two for the result to be even. You write down the first
combination of numbers: . Next, if the spinner lands on
, it can land on any number on the second
spinner. We now have the combinations of and
. Finally, if the first spinner ends on
, we
have Since there are
possible combinations, and we have
evens, the final answer is
.
Solution 3
We can also list out the numbers. Box A has chips ,
, and
, and Box B also has chips
,
, and
. Chip
(from Box A)
could be with 3 partners from Box B. This is also the same for chips and
from Box A.
total sums. Chip
could be
multiplied with 2 other chips to make an even product, just like chip
. Chip
can only multiply with 1 chip.
. The answer is
.
Solution 4
Here is another way:
Let's start by finding the denominator: Total choices.
There are chips we can choose from in the 1st box, and
chips we can choose from in the 2nd box. We do
, and get
.
Now - to find the numerator: Desired choices.
To get an even number, we need to pick 2 from at least one of the boxes. There are
choices as to finding which box we will draw the 2 from. Then, we have
choices from the other box to pick any of the other chips,
and
.
However, we are over counting the configuration twice, and so, we subtract that one configuration from our total.
.
Thus, our answer is
.
~ del-math.
Solution 5
This might take longer to solve, but you definitely will have the right answer.
You first list out all the possible combinations regardless if the product is even or odd.
There are 9 possible combinations, and
of the combinations’ products are even. So, our answer is
.
~MiracleMaths
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.