Question

A pair of numbers is picked up randomly (without replacement) from the set $[1,2,3,5,7,11,12,13,17,19]$. The probability that the number $11$ was picked given that the sum of the numbers was even is nearly:

• A. $0.1$
• B. $0.125$
• C. $0.24$
• D. $0.18$

Answer 1

Answer: (C)

$0.24$

Let $A$ be the event such that the number $11$ is picked and $B$ be the event such that the sum of the numbers is even.

We know that, sum of two numbers will be even only when both numbers are odd or both numbers of even.

Therefore, $n\left(B\right)=2C_2+8C_2=1+28=29$

Let $A\cap B$ be such that the number $11$ is chosen and the sum is even.

Since $11$ is chosen, to make the sum even we should choose another odd number. Apart from $11$, there are $7$ odd numbers that can be chosen to make the sum even.

$\therefore n(A\cap B)=7C_1=7$

Now we have to find the probability that the number $11$was picked such that the sum of the numbers is even.

That is, we have to find $P\left(A/B\right)$

We know that $P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$

Since $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)},P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}$

Thus we get $P\left(A/B\right)=\frac{\frac{n(A\cap B)}{n\left(S\right)}}{\frac{n\left(B\right)}{n\left(S\right)}}$

$\Rightarrow P\left(A/B\right)=\frac{n(A\cap B)}{n\left(B\right)}$ $=\frac{7}{29}$

$\Rightarrow P\left(A/B\right)=0.24$

Therefore, the probability that the number $11$was picked such that the sum of the numbers was even is $0.24$.