Probability Ques 80
Two integers are selected at random from the set $1,2$,$\ldots . . ., 11$. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is
(2019 Main, 11 Jan I)
(a) $\frac{2}{5}$
(b) $\frac{1}{2}$
(c) $\frac{7}{10}$
(d) $\frac{3}{5}$
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Answer:
Correct Answer: 80.(a)
Solution:
Formula:
- In $\{1,2,3, \ldots ., 11\}$ 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. Then, $n(A)={ }^{5} C _2$ and $n(B)={ }^{5} C _2+{ }^{6} C _2$
Required probability
$ \begin{aligned} P(A / B)=\frac{P(A \cap B)}{P(B)} & =\frac{{ }^{5} C _2 /{ }^{11} C _2}{\frac{\left({ }^{6} C _2+{ }^{5} C _2\right)}{{ }^{11} C _2}} \\ & =\frac{{ }^{5} C _2}{{ }^{6} C _2+{ }^{5} C _2}=\frac{10}{15+10}=\frac{2}{5} \end{aligned} $
BETA
