Question

Two integers are selected at random from the set {1, 2, …., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is:
1. 7/10
2. 1/2
3. 2/5
4. 3/5

Answer 1

Correct Answer - Option 3 : 2/5

In {1, 2, 3, …, 11} there are 5 even numbers and 6 odd numbers. The sum even is possible only when both are odd or both are even.

Let A be the event that denotes both numbers are even and B be the event that denotes sum of numbers is even.

Now,

⇒ n(A) = ⁵C₂ and n(B) = ⁵C₂ + ⁶C₂

Now, the required probability is:

\(\Rightarrow P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\) 

\(\Rightarrow P\left( {\frac{A}{B}} \right) = \frac{{\frac{{{}_\;^5{C_2}}}{{{}_\;^{11}{C_2}}}}}{{\frac{{\left( {{}_\;^6{C_2}{ + ^5}{C_2}} \right)}}{{11{C_2}}}}}\)

\(\Rightarrow P\left( {\frac{A}{B}} \right) = \frac{{{}_\;^5{C_2}}}{{{}_\;^6{C_2}{ + ^5}{C_2}}}\) 

\(\Rightarrow P\left( {\frac{A}{B}} \right) = \frac{{10}}{{15 + 10}}\) 

\(\therefore P\left( {\frac{A}{B}} \right) = \frac{2}{5}\)