arjunshn wrote:A set P = {1, 2, 3, 4, 5} and set Q = {1, 2, 3, 4, 5, 6, 7} is defined with given numbers. Now one number is chosen randomly from each of the given set in such a way that the absolute difference between the two numbers is 2. What is the probability that one of the number chosen is 3 ?

Please post the exact wording of the original question along with the source. On this particular question we can determine what's being asked, but grammatical errors (of which there are a few in this post) can alter the meaning of a question. Also, please post the answers, since many GMAT strategies revolve around using the choices.

Let's start solving this question by noting the probability formula (on test day you should always jot applicable formulae down on your noteboard):

Probability = (# of desired outcomes)/(total # of possibilities)

The denominator is the easier part of the fraction to find, so let's start by calculating the total number of ways we can get two numbers that are 2 apart. I'm sure there are fancy methods, but brute force is probably the quickest.

Focusing on set P, since that's smaller:

1 in P goes with 3 in Q.
2 in P goes with 4 in Q.
3 in P goes with 1 or 5 in Q.
4 in P goes with 2 or 6 in Q.
5 in P goes with 3 or 7 in Q.

So, there are 8 pairs we can select that will have an absolute difference of 2.

For the number of desired outcomes, we now simply count the possibilities than include 3:

1 in P with 3 in Q.
3 in P with 1 or 5 in Q.
5 in P with 3 in Q.

So, there are 4 pairs that match our desired outcome.

Finally, we plug into the equation:

Probability = 4/8 = 1/2... all done!