OA: [spoiler](D)[/spoiler]

Anurag@Gurome wrote:

tabsang wrote:Is the probability of an element in Set B also being an element in Set A equal to the probability of an element in Set A also being an element in Set B?

(1) Set A consists of all integers from 1 through 5 and Set B consists of all prime numbers less than 10.

(2) Set A consists of all odd numbers between 1 and 10, both inclusive, and Set B consists of all even numbers between 1 and 10, both inclusive.

As for both statements, we can definitely determine all the elements of set A and set B, we can definitely calculate the required probabilities. Hence, we can compare them and answer the question with definite YES or NO.

Hence, both statements are individually sufficient to answer the question.

The correct answer is D.

Thanks Anurag

Here's my approach but I'm not sure if my working is correct (from a conceptual point of view).
I too got [spoiler](D)[/spoiler] as the answer.

According to Statement 1:
A={1,2,3,4,5} = 5
B={2,3,5,7} = 4
A∩B={2,3,5} = 3

Now, P(element in Set B also being an element in set A) = P(B|A) .....(is this correct?)
P(B|A)=P(A∩B)/P(A)=3/5

Also, P(element in Set A also being an element in set B) = P(A|B) .....(is this correct too?)
P(A|B)=P(A∩B)/P(B)=3/4

SUFFICIENT.

According to Statement 2:
A={1,3,5,7,9} = 5
B={2,4,6,8,10} = 5
A∩B={} = 0

Now, P(element in Set B also being an element in set A) = P(B|A) .....(correct?)
P(B|A)=P(A∩B)/P(A)=0

Also, P(element in Set A also being an element in set B) = P(A|B) .....(correct?)
P(A|B)=P(A∩B)/P(B)=0

SUFFICIENT.

So you see, although I've arrived at [spoiler](D)[/spoiler] (didn't really need to do the calculations), I'm not sure if I've correlated the concept and the question correctly (i.e P(element in Set B also being an element in set A) = P(B|A))

Cheers,
Taz

OA IS D