Set M is comprised of the first 5 positive integer multiples of 3 and

Question

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

A. 2/5
B. 7/15
C. 8/15
D. 5/8
E. 2/3

quantumliner · Answer

As per question

M = {3,6,9,12,15}
N = {1,2,3}

Total Number of ways to select 1 number from each set = 5*3 = 15

Total number of ways a number is selected at random from each set, such that sum will be odd =

Total number of ways of selecting odd numbers from Set M and even numbers from Set N + Total number of ways of selecting even numbers from Set M and odd numbers from Set N

Total number of ways of selecting odd numbers from Set M and even numbers from Set N = 3*1 = 3
Total number of ways of selecting even numbers from Set M and odd numbers from Set N = 2*2 = 4

Total number of ways a number is selected at random from each set, such that sum will be odd = 3+4=7

Probability that that sum will be odd, If a number is selected at random from each set = 7/15

Answer is B. 7/15

gmatexam439 · Answer

Ans - B

M={3,6,9,12,15}
N={1,2,3}

Total possible outcomes = 5C1*3C1 = 15 ........(A)
Total favourable outcomes = 3*1+2*2 = 7 ........(B)
Therefore, probability = A/B = 7/15

In (B), the sum is always an odd number when we add Even and Odd number (E+E=E; O+O=E)
Thus, in set M, there are 3 ways to select Odd number and in set N only 1 to select even number
Similarly, in set N, there are 2 ways to select Odd number and in set M 2 ways to select even number

0akshay0 · Answer

M = {3,6,9,12,15} and N = {1,2,3} Total possibilities =5*3 =15 Favorable possibilities = (1,4) (1,12) (2,3) (2,9) (2,15) (3,6) (3,12) = 7 probability that that sum will be odd = 7/15 Hence option B is correct Hit Kudos if you liked it 8-)

ScottTargetTestPrep · Answer

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

shashankism · Answer

Set M = first 5 positive integer multiples of 3 = {3,6,9,12,15}
Set N = first 3 positive integer multiples of 1 = {1,2,3}

So total no. of possible set = 5 * 3 = 15

2 cases of set with sum of the nos. as odd is possible 
case 1 : Set M {3,9,15} & Set N {2} = 3*1 =3
case 2 : Set M {6,12} & Set N {1,3} = 2 * 2 = 4

So, total number = 3+4 = 7

Probability (sum is odd) = 7/15

Answer B

mbaaspirant80 · Answer

Set M = first 5 positive integer multiples of 3 = {3,6,9,12,15}
Set N = first 3 positive integer multiples of 1 = {1,2,3}

So total no. of possible set = 5 * 3 = 15

odd is possible only when one no is odd and another is even: even +odd = odd;
case 1 : when we select odd from Set M {3,9,15} & even from Set N {2} = 3*1 =3
case 2 : when we select odd from Set N {1,3} & even from Set M {6,12} = 2 * 2 = 4

So, total number = 3+4 = 7

Probability that sum is odd = likely outcome/total outcome=7/15

Answer B

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Set M is comprised of the first 5 positive integer multiples of 3 and

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