GMAT / GRE Tutorial Series: Probability | Theory, Solved Examples & Practice Questions

GMAT Probability Questions

MS and MBA applicants often ask questions with regard to the GMAT probability Questions of their chances of getting into top colleges or getting a scholarship. As these are dependent on several factors, it’s impossible to give an accurate answer.

Understanding the basic rules and formulas of probability will help you score well in your GRE / GMAT exams. 

Also Check GMAT / GRE Tutorial Series Set Theory

Probability: Meaning and definition

According to the Oxford dictionary, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’.

In the field of mathematics, probability is a numerical description of the likelihood of the occurrence of an event.

The following are examples of events:

  • Tossing a coin with heads as the outcome
  • Drawing a red pencil from a set of different colored pencils
  • Drawing a card from a deck of 52 cards etc
  • Rolling dice to get a particular number

Either an event will occur for sure, or it will not occur at all. Or there are different levels of possibilities that the event may occur.

An event that is sure to occur is called a Certain event and its probability is 1.

An event that never happens at all is called an Impossible event and its probability is 0.

This means that all other possibilities of an event occurrence lie between 0 and 1.

This is shown as follows:

0 <= P(A) <= 1

where A is an event and P(A) is the probability of the occurrence of the event.

This also implies that a probability value can never be negative or less than zero.

The ‘sample space’ is a set of all the possible outcomes of an experiment.

Take for example the tossing a coin.

When a coin is tossed, the possible outcomes are Head and Tail. So, the sample space is represented as {H, T}.

Similarly when two coins are tossed, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

The probability of getting a head each time you toss the coin is 1/2. So is the probability for getting a tail.

Also Check GMAT / GRE Tutorial Series Accounting

The Basic Formula of Probability

As you already familiar with the list of GMAT maths formulas, the Probability of the occurrence of an event A is defined as:

P(A) = (No. of ways A can occur)/(Total no. of possible outcomes)

Another example is the rolling of dice. When a single die is rolled, the sample space is {1,2,3,4,5,6}.

What is the probability of rolling a 5 when a die is rolled?

No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = 1/6.

Compound probability

Compound probability relates to the likelihood of the occurrence of two independent events.

The formula for the compound probability

P(A or B) = P(A) + P(B) – P(A and B)

where A and B are any two events.

P(A or B) is the probability of the occurrence of at least one of the events.

P(A and B) is the probability of the occurrence of both events, A and B at the same time.

Also Check GMAT / GRE Tutorial Series Probability

Mutually Exclusive Events

In mutually exclusive events, the occurrence of one event indicates the non-occurrence of the other event

OR

When two events cannot occur at the same time, they are considered mutually exclusive.

Note: For a mutually exclusive event, P(A and B) = 0.

Example 1: What is the probability of getting a 2 or a 5 when a die is rolled?

Solution:

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,

Probability of getting a 2 or a 5,

P(2 or 5) = P(2) + P(5) – P(2 and 5)

==>      1/6 + 1/6 – 0

==>      2/6 = 1/3.

Example 2: Consider the example of finding the probability of selecting a red card or a 4 from a deck of 52 cards.

Solution:

We need to find out P(R or 4)

Probability of selecting a red card  = 26/52

Probability of selecting a 4 = 4/52

Probability of selecting both a red card and a 4 = 2/52

P(R or 4)= P(R) + P(4) – P(R and 4)

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13

Independent and Dependent Events

Independent Event

In the case of multiple events that happen, when the outcome of one event DOES NOT affect the outcome of the other events, they are called independent events.

Consider a die is rolled twice. The outcome of the first roll doesn’t affect the second outcome. These two are independent events.

Example 1: If a coin is tossed twice, what is the probability of getting two consecutive tails?

Probability of getting a tail in one toss = 1/2

The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer.

Here’s the verification of the above answer with the help of sample space.

When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

Example 2: Consider another example where a box contains 4 blue, 2 red and 3 black socks. If a sock is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue socks and 1 black sock?

Solution

Here, total number of socks = 9

Probability of drawing 1 blue sock = 4/9
Probability of drawing another blue sock = 4/9
Probability of drawing 1 black sock = 3/9
Probability of drawing 2 blue socks and 1 black sock = 4/9 * 4/9 * 3/9 = 48/729 = 16/243

Dependent Events

When the outcome of one event affects the outcome of another event, they are called dependent events.

Consider the same example of drawing a sock from a box, but with a slight difference.

Example 1: A box contains 4 blue, 2 red and 3 black socks. If 2 socks are drawn at random from the box, NOT replaced and then another sock is drawn. What is the probability of drawing 2 blue socks and 1 black sock?

Solution:

Probability of drawing 1 blue sock = 4/9
Probability of drawing another blue sock = 3/8
Probability of drawing 1 black sock = 3/7
Probability of drawing 2 blue socks and 1 black sock = 4/9 * 3/8 * 3/7 = 1/14

Here’s another one to better understand this concept.

Example 2: What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement?

Probability of drawing a jack = 4/52 = 1/13

After drawing one card, the number of cards is 51.

Probability of drawing a queen = 4/51.

Now, the probability of drawing a jack and queen consecutively is 1/13 * 4/51 = 4/663

Conditional probability

Conditional probability is calculating the probability of an event occurring given that another event has already occurred.

The formula for conditional probability P(A|B), read as P(A given B) is

P(A|B) = P (A and B) / P(B)

Consider the following example:

Example: In a class, 40% of the students study biology and chemistry. 60% of the students study biology. What is the probability of a student studying chemistry given he/she is already studying biology?

Solution

P(B and C) = 0.40

P(B) = 0.60

P(C|B) = P(B and C)/P(C) = 0.40/0.60 = 2/3 = 0.67

Complement of an event

A complement of an event A is when there is NO occurrence of event A.

A complement of an event is denoted as P(Ac) or P(A’).

P(Ac) = 1 – P(A) Or  P(A)+P(Ac) = 1

For example,

if W is the event of getting a head in a coin toss, Wc is not getting a head i.e., getting a tail.

if W is the event of getting an even number in a die roll, Wc is the event of NOT getting an even number i.e., getting an odd number.

if W is the event of randomly choosing a number in the range of -3 to 3, Wc is the event of choosing every number that is NOT negative i.e., 0,1,2 & 3 (0 is neither positive or negative).

Here’s another example:

Example: A single coin is tossed 5 times. What is the probability of getting at least one head?

Solution:

Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Sample Probability questions with solutions

Example 1

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled?

Solution

Let the event of the occurrence of a number that is odd be ‘M’ and the event of the occurrence of a number that is less than 5 be ‘N’. We need to find P(M or N).

P(M) = 3/6 (odd numbers = 1,3 and 5)

P(N) = 4/6 (numbers less than 5 = 1,2,3 and 4)

P(M and N) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)

Now, P(M or N)            = P(M) + P(N) – P(M or N)

= 3/6 + 4/6 – 2/6

P(M or N) = 5/6.

Example 2

A box contains 4 chocolates and 4 candies. Tony eats 3 of them, by randomly choosing. What is the probability of choosing 2 chocolates and 1 candy?

Solution

Probability of choosing 1 chocolate = 4/8 = 1/2

After taking out 1 chocolate, the total number is 7.

Probability of choosing 2nd chocolate = 3/7

Probability of choosing 1 candy out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocolate and 1 candy = 1/2 * 3/7 * 2/3 = 1/7

Example 3

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

Solution

Let the event of getting a greater number on the first die be R.

There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.

And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, R = {(5,3), (6,2)}.

Therefore, P(Sum equals 8) = 5/36 and P(R) = 2/36.

Now, P(R|sum equals 8)= P(R and sum equals 8)/P(sum equals 8)

= (2/36)/(5/36)

= 2/5

Probability Quiz: Sample probability questions for practice

1) A bag contains blue and red pencils. Two pencils are drawn randomly without replacement. The probability of selecting a blue and then a red pencil is 0.2. The probability of selecting a blue pencil in the first draw is 0.5. What is the probability of drawing a red pencil, given that the first pencil-drawn was blue?
a) 0.4
b) 0.2
c) 0.1
d) 0.5

Answer A.

2) A die is rolled thrice. What is the probability that the sum of the rolls is at least 5.
a) 1/216
b) 1/6
c) 3/216
d) 212/216

Answer D.

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