For most students, one of the most dreaded topics in the GMAT Quant section is Probability. Many students struggle with this topic and hence are seen preparing vigorously. Having said that, students need not worry too much about this topic. The reason being, one, most often only 1 or 2 questions are asked from this topic. Being asked more than 2 questions is extremely rare. Second, even if you get a probability question, it doesn’t necessarily have anything to do with the formula or properties. Questions mostly test logic and reasoning skills, hence, as long as your basics are clear, you can easily solve the GMAT quant arithmetic probability questions.

Moreover, to solve probability questions, it is very essential to know the concepts of permutations and combinations. In this article, we will discuss the basics of probability and how to solve questions from this topic. To begin with, let’s first tell you what probability is.

Probability is the extent to which an event is likely to occur, measured by the ratio of the favorable cases to the total number of cases possible.

That is,

Probability = Favorable Outcomes/Total Outcomes

GMAT Quant Overlapping & Non-overlapping Sets questions

GMAT Quant Overlapping & Non-overlapping Sets solutions

GMAT Quant Arithmetic Probability

We will discuss the approach to solve a Probability question, but before that some key points to note.

1. Sample space: It is a collection of all possible outcomes of a random experiment.

2. PEvent happens+PEvent doesn’t happen=1

3. 0≤Probability≤1

You should know the sample space of three events which are commonly tested in the GMAT exam.

Tossing a coin:

If coin is tossed n times, then possible number of outcomes = 2n

Rolling a dice:

If dice is rolled n times, then possible number of outcomes = 6n

ack of cards:

In a deck of 52 cards, we have 4 suits: There are 26 red cards and 26 black cards.

In 26 red cards we have 13 cards of Heart suit and 13 of Diamond suit and in 26 black cards we again have 13 cards of Spade and 13 of Club.

In a particular suit (each suit) we have 4 special cards: King(K), Queen(Q), Jack(J), 1 Ace and 9 numbered cards from 2 to 10.

Now, let’s come to the important part, that is, the approach of solving a “Probability question”.

Ways of Solving a Probability Question

There are two ways to solve a probability question:

1. Combinatorial Approach

Using combinations (nCr=n!r!*n-r!) to select the total number of outcomes and favorable number of outcomes.

2. Complex- event Approach

Probability of Complex Event = Product of Probability of Individual Events * Arrangement

Probability Sample Questions

A bag contains 2 red, 3 green and 4 yellow balls. Two balls are selected simultaneously. What is the probability of getting at-least one red ball?

Let’s understand the information given in the question.

Total = 9 balls

Red = 2 and non-Red = 7

At least translates to ≥ in math

The probability of selecting more than one red ball, when two balls are selected simultaneously —

So, it could be either 1 red ball and 1 non-red ball OR it could be both RED.

Hence, we need to find

Probability of (One Red and One Non-Red) + Probability of (Both Red)

1. Combinatorial Approach:

Probability of (One Red and One Non-Red) = One red * one non-redSelecting two balls from total 9

Selecting one red ball = 2C1=2

Selecting one non-red ball = 7C1=7

Selecting two balls from total balls = 9C2=36

Probability of (One Red and One Non-Red) = 1436

Now,

Probability of (Both Red) = Two red ballsSelecting two balls from total 9

Selecting two red balls = 2C2=1

Selecting two balls from total balls = 9C2=36

Probability of (Both Red) = 136

Probability of (One Red and One Non-Red) + Probability of (Both Red) = 1436+136=1536=512

2. Complex-event Approach:

Probability of (One Red and One Non-Red) = P (Red) * P (Non-Red) * Arrangement.

P(Red) = No. of red ballstotal number of balls=29

P(Non-Red) = No. of non red ballsRemaining total number of balls=78 (Why only 8 in the denominator? As one red ball picked first, then the number of red balls in the second pick will reduce by 1).

Arrangements, here there are two different events, P(Red) and P(Non-red) two events can be arranged in 2! Ways.

So, Probability of (One Red and One Non-Red) = 29*78*2!=1436

Probability of (Both Red) = P (Red) * P (Red) * Arrangement

P(Red) = No. of red ballstotal number of balls=29

P(Red) = Remaining No. of red ballsRemaining total number of balls=18 (As one red ball picked first, then the number of red balls in the second pick will reduce by 1, so the numerator and denominator reduces by 1).

Arrangements — here are two events but both are identical, P(Red) and P(Red) two identical events can be arranged in only 1 way.

Probability of (Both Red) = 29 * 18*1 = 136

Probability of (One Red and One Non-Red) + Probability of (Both Red) = 1436+136=1536=512

Alternatively, we can solve this using “reverse” strategy, that is,

PEvent happens+PEvent doesn’t happen=1

Pat least one red+PNo red=1

Pat least one red=1-PNo red

Combination approach:

PNo red = two non-red ballsSelecting two balls from total 9=7C29C2=2136

Pat least one red=1-PNo red=1-2136=1536=512

Complex-event approach:

Probability of (Both non-red) = P (Non-Red) * P (Non-Red) * Arrangement

Probability of (Both non-red) = 79*68*1=2136

Hence,
Pat least one red=1-PNo red=1-2136=1536=512

It is very important to have a clear approach when solving a Probability question. You can use any of the above approaches. Remember only if the approach is clear, then you would be able to address the other aspects of the Probability questions.

Other Aspects of the Probability Questions

Another confusion most students have in these topics is when do they add and when do they  multiply.

Whenever there is the  word “And” given, you should  ‘multiply’ the probability.

If two or more events happen simultaneously, then we need to look at whether they are independent events or dependent events.

In general, if the words “replaced” “put back” or any similar words are mentioned in the question, then the events are independent.

If those words are not mentioned in the question, then the events are dependent.

Whenever there is a word “OR” given, you should ‘add’ the probability.

If any of the events happen, that is, event A or event B, then we need to add the probabilities.

There are two-types here,

• Add the probabilities, if the events are mutually exclusive(no-overlap), then

PA or B=PA+PB

• If the events are non-mutually exclusive (overlap between two events), then we need to subtract the probability of both happening together after adding the probabilities.

PA or B=PA+PB-P(A and B)

Let’s look at some examples:

1. There is a 10% chance that it won’t snow during winter, there is a 20% chance that the workplace will not be closed during winter. What is the probability that it will snow and the workplace will be closed during the winter?

Probability of it will snow = 1 – Probability it won’t snow) =1 – 0.1 = 0.9

Probability of workplace will be closed = 1 – Probability of workplace will be open = 1 – 0.2 = 0.8

Since we can see the word “AND” in the question we need to multiply the probability.

Hence, the probability is 0.9*0.8=0.72

2. What is the probability of selecting an integer randomly between 1 to 150(both inclusive) which is a multiple of a 5 OR multiple of a 7?

Since there is a word “OR” in the question, we need to add and look if there is an overlap or not.

PA or B=PA+PB-P(A and B)

Probability of selecting a number which is multiple of 5 or 7?

Pmultiple of 5 or 7

=Pmultiple of 5+Pmultiple of 7-P(multiple of 5 and 7)

Total number of integers between 1 to 150(inclusive) = 150

Multiples of 5 between 1 to 150 = 5, 10, …. 150 = 30 multiples

Probability of selecting a multiple of 5 = 30150

Multiples of 7 between 1 to 150 = 7, 14, …. 147 = 21 multiples

Probability of selecting a multiple of 7 = 21150

There are some multiples of both 5 and 7 between 1 to 150, i.e., multiple of 35 = 35, 70, 105 and 150 = 4 multiples

Probability of selecting both multiple of 5 and 7 = 4150