# GRE Quantitative Data Analysis – Probability

Usually, you won’t be facing many probability questions in the GRE exam, and this is why many test aspirants start prioritizing questions that involve arithmetic algebra and geometry. But, if you’ve mastered the concepts of these topics, then turning your attention to probability can boost your performance on the D-day. Just like all the other concepts of GRE, the question of probability is also based on the basic concepts; the nuts, and bolts of probability that you studied in your grade school. However, to help you revise, we’ll guide you through the basic concepts of probability to master it on the D-day.

Formula of Probability

Probability asks you to find the likelihood of occurrence of an event. With this, it is easy to comprehend that the formula of probability is really straight forward. The range of an event happening always lies in between zero and one. This means that probability of an event can neither be negative nor more than one.

The probability of happening of an event is denoted by P (E), whereas, the probability of not happening an event is denoted by P (E’).

$Probability \; of \; the \; Occurrence \; of \; an \; event, P(E) = \frac{Number \; of \; Favourable \; outcomes (E)}{Total \; outcomes (S)}$

Types of Events:

Basically, there are two types of events:

1. AND event: In this case unless and until all the conditions hold true, it will not be added in the sets of events
2. OR event: In this case if even a single condition holds true it will be included in the set of events

Let’s solve a problem to understand the working of this formula.

Question: A bag contains tickets numbered from 1 to 20. If a ticket is drawn at random from the bag then what is the probability that the ticket drawn from the bag has a number which is a multiple of 2 or 3?

• $A) \; \frac{1}{2}$
• $B) \; \frac{13}{20}$
• $C) \; \frac{2}{5}$
• $D) \; \frac{7}{20}$

Solution: In this question; Sample Space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

Let event, E = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}

$Probability, P(E) = \frac{13}{20}$ Answer is B.

This question is far easier than what you will actually face on GRE exam day but having your basic concepts concrete allows you to solve all complex questions with great ease.

Let us solve another question but from the practice sets of GRE.

Question: A is chosen randomly from a set of {2, 5, 7, 16, 21}
B is chosen randomly from the set of {6, 11, 19, 22, 23, 27}
Calculate the probability that the sum of A and B will be 27.

Solution: Since, any of the first set can be combined with the second. Hence, total combinations = 5 ×6=30
Out of these combinations, only (5 + 22), (17 + 11), and (21 + 6) makes 27
So, P (E) = 330 = 110=0.1

## Conditional Probability

Conditional Probability describes a certain type of question which arises in Probability. In probability, there can be two types of events: Independent and Dependent events.

• ### Independent Event

An event can be described as an Independent event if that particular event is not affected by other events. For example, let us take the question of the probability of getting two heads in a row on flipping of a coin. Each tossing of the coin is an isolated event, and it is not influenced by the previous actions. With every throw, there is a 50% chance of getting a head.

• ### Dependent Event

An event can be described as a Dependent event if the event can be affected by any previous event. For example, let us take 5 stones, of which 2 are red and 3 are green, in a bag. You need to take 2 stones at random from the bad. What is the probability that both the balls are red?
In this question, there are two separate events. First is to pick the first stone and the second event is to pick the second stone.

Normally, your answer would be ⅖ . But after picking the first stone, the chances changes. If you pick a green stone at the first attempt, the probability of getting the red stone changes to ½ (2/4). If you pick a red stone at the first attempt, then the probability of getting the red stone in the second event changes to ¼ .

This is the example of conditional probability, in which the result of the previous event comes into consideration.

Bayes Theorem

P(A) = Probability of event A

P(B) = Probability of event B

$P(A|B)=\frac{P( B|A ) P( A ) }{ P( B|A ) P( A ) P ( B|A )P( A ) + P( B|A) P( A ) }$

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