Bunuel wrote:
Set M is comprised of the first 5 positive integer multiples of 3 and set N is composed of the first 3 positive integer multiples of 1. If a number is selected at random from each set, what is the probability that that sum will be odd?
A. 2/5
B. 7/15
C. 8/15
D. 5/8
E. 2/3
Set M contains the values of 3, 6, 9, 12, and 15. Set N contains the values 1, 2, and 3.
We need to determine the probability when a number is selected from each set that the sum is odd.
To get an odd sum, we can have even + odd or odd + even, that is, an even number from M and an odd number from N OR an odd number from M and an even number from N.
Let’s start by calculating the probability of obtaining an even number from M and an odd number from N:
P(even number from set M) = 2/5
P(odd number from set N) = 2/3
Thus, the probability of an even number from M and an odd number from N is 2/5 x 2/3 = 4/15.
Next let’s determine the probability of an odd number from M and an even number from N:
P(odd number from set M) = 3/5
P(even number from set N) = 1/3
Thus, the probability of an odd number from M and an even number from N is 3/5 x 1/3 = 3/15.
Thus, the probability of an odd sum is 4/15 + 3/15 = 7/15.
Answer: B
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