Two integers are selected at random from the first 11 natural numbers. If the sum of the integers is even, then the probability that both the numbers are odd is:

Correct option is B

Given:

First 11 natural numbers

Formula Used:

Probabilities and Combination

Probability P(A$$) =\frac{Number\ of \ Favourable\ Outcomes }{Total\ Number \ of \ Outcomes}$$​​

​$$^nC_r = \frac{n!}{r!(n-r)!}$$​​

Solution:

First we need to determine the total number of ways to choose 2 numbers out of 11 using combination formula:

Odd integers = 1, 3, 5, 7, 9, 11

Even integers = 2,4, 6, 8, 10

For sum of two integers to be even the numbers either should be both even or both odd

Ways to select two numbers to be odd is $$^6C_2 = \frac{6!}{2!(6-2)!} = 15$$​​

Ways to select two numbers to be even is $$^5C_2 = \frac{5!}{2!(5-2)!} = 10$$​​

Total outcomes that sum is even is 15 + 10 = 25

Total outcomes that integers selected are odd = 15

Now probability is =$$\frac{15}{25} = \frac{3}{5}$$​​