Difference between revisions of "2015 AMC 8 Problems/Problem 7"
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3 x 3 = 9 | 3 x 3 = 9 | ||
There are 9 possible combinations, and 5 of the combinations’s products are even. So, our answer is our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | There are 9 possible combinations, and 5 of the combinations’s products are even. So, our answer is our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
+ | - MiracleMaths | ||
==See Also== | ==See Also== |
Revision as of 14:20, 20 December 2021
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
We can instead calculate the probability that their product is odd, and subtract this from . In order to get an odd product, we have to draw an odd number from each box. We have a
probability of drawing an odd number from one box, so there is a
probability of having an odd product. Thus, there is a
probability of having an even product.
Solution 2
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
, to be even, it must land on two. 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 . 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. Hat A has chips ,
, and
, and Hat B also has chips
,
, and
. Chip
(from Hat A)
could be with 3 partners from Hat B. This is also the same for chips and
from Hat 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,
.
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.
1 x 1 = 1
1 x 2 = 2
1 x 3 = 3
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
There are 9 possible combinations, and 5 of the combinations’s products are even. So, our answer is our answer is .
- MiracleMaths
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.