Question

A = {1,2,3.... 11} two numbers are selected and sum is even. Then the conditional probability that both numbers are even is?
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Answer 1

Answer:

2/11

Step-by-step explanation:

favourable outcomes/ total number of outcomes

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Here

$n(A) = 11$

Since two numbers are selected

So the total number of possible outcomes = 121

Let E = The set of even numbers in A

$= \{2,4,6,8,10 \}$

So

$n(E ) = 5$

Let H be the event that two numbers are selected and sum is even

Then total number of possible outcomes for the event H is

$= (6 \times 6 + 5 \times 5) = 61$

So

$\displaystyle \: P(H) = \frac{61}{121}$

Again let D be the event that two numbers are selected and both numbers are even

Then the total number of possible outcomes for the event D is

$= 5 \times 5 = 25$

So

$\displaystyle \: P(D) = \frac{25}{121}$

$H \cap D$

is the event that two numbers are selected, sum is even and both numbers are even

$\displaystyle \: P(H \cap D \: ) = \frac{25}{121}$

So the required conditional probability

$\displaystyle \: P( \frac{D}{H} )$

$\displaystyle \: = \frac{P( D \cap \: H )}{P( H )}$

$= \displaystyle \: \frac{ \frac{25}{121} }{ \frac{61}{121} }$

$= \displaystyle \: \frac{25}{61}$