How to Solve GMAT Probability & Number Properties Question using Sets

GMAT Sets and Number PropertiesWhen you think about Set Questions, the traditional form of representing numbers in Sets comes to our mind. In GMAT set questions however, several other concepts will be tested simultaneously. Let us look into an example where Set representation is used to  solve Probability and Number properties GMAT Question.

Q) Rhonda’s Chocolate factory is creating packets of chocolates with 12, 13, 14, 25, 35, 44, 66, 77 and 88 chocolates in each packet. The manager at the Factory arranged the packets in such a way that all the bright colored packets were in one group, and dark colored packets in other. If the bright colored packet group had packets with 12, 25, 77, and 88 chocolates, and dark colored packet group had the remaining, what is the probability that picking a pair from dark and bright colored packet group gives even number of chocolates?


a)    1/3
b)    1/2
c)    1/5
d)    2/3
e)    1/8


The length of the question should not intimidate you. This simple set question tests the theory of Number properties.

First, let us understand the information in the question.

There are two packet groups – light and dark colored.

The bright colored packet group has chocolate packets with 12, 25, 77, and 88 chocolates. Let us represent this as a set A.

A = {12, 25, 77, 88}

The dark colored packet group had the remaining chocolates. Let the set be represented by B

B = {13, 14, 35, 44, 66}

The question is what is the probability that picking a pair from dark and bright colored packet group gives even number of chocolates?

Total Number of pairs that can be created from A and B = 4 x 5 = 20

Total Number of ways in which packets can be picked from A & B such that the sum is even = n

Probability = n/20

The only possibility of getting even numbers is when two packets picked from A & B are both even or each packets are odd

Even + Even = Even
Odd + Odd = Even

Let us pick each number from A

(12,14) (12,44) (12,66)
(25,13) (25,35)
(77, 13) (77, 35)
(88, 14) (88, 44) (88, 66)

Total (n) = 10

Probability = 10/20 = 1/2


A = {12, 25, 77, 88}

From A, when you pick the first number – 12 it is an even number

For even number 12, when you pick another number from B that satisfy the even sum condition, you will get 3 groups

For even number 88, the number of groups would be the same -> 3

For odd number 25, when you pick another number from B that satisfy the even sum condition you will get 2 groups

For odd number 77, the number of groups would be the same -> 2

Total (n) = 3*2 + 2*2 = 10

Correct Answer: B

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