Probability is one of the most important topics of quantitative aptitude section of CAT exam, it tests candidates mathematical as well as reasoning skills. To help candidates with probability here we are providing some important definition related to probability which will help a candidate to learn this topic from the scratch for CAT examination.

**Sample Space in Probability:**

#### A sample space is the set of all possible outcomes of a particular experiment.

Consider the experiment of flipping two coins The sample space will be {HH, HT, TH, TT }

where H → Head and T→ Tail

An event is a subset of the sample space.

For example, the event that a head shows up will be {HH, HT, TH}

Also, the sum of all the probabilities in the sample space is 1.

**The Complement of an Event:**

#### If E is an event within the sample space S of an activity or experiment, the complement of E (denoted E’) consists of all outcomes in S that are not in E. The complement of E is everything else in the problem that is NOT in E.

**Example:**

**Experiment: Tossing a coin **

Event E The coin shows heads.

Complement E’ The coin shows tails.

The probability of the complement of an event is one minus the probability of the event.

**⇒ P(E’) = 1 – P(E).**

**Probability: Illustrations**

**i) A pair of dice is rolled. What is the probability of not rolling doubles? **

Sample space= rolling of two dies = 6 × 6= 36 Doubles are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) P(doubles) = 6/36 = 1/6 P(not doubles) = 1 – 1/6 = 5/6 e.g.

So, according to the law of probability, P(not doubles) = 1 – 1/6 = 5/6 e.g.

**ii) A pair of dice is rolled. What is the probability of rolling 9 or less? **

Sample space= rolling of two dies = 6 × 6 = 36 The complement of rolling “9 or less” is rolling 10, 11 or 12. P(9 or less) = 1 – P(10, 11 or 12) = 1 – [P(10) + P(11) + P(12)] (to get a 10 there are 3 ways (6, 4),(4, 6) and (5, 5) and to get a 11, there are 2 ways (6,5) and (5, 6) & to get a 12, there is only one way (6, 6)) = 1 – (3/36 + 2/36 + 1/36) = 5/36.

**Mutually exclusive events:**

Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common)

**Example: **

**What’s the probability of getting either heads or tails when flipping a coin once? **

**Solution:**

Since the only possible outcomes are heads or tails, we expect the probability to be 100%, or 1/2 : 1/2 = 1.

The events Heads and tails are mutually exclusive. That is, if heads occurs, then tails cannot (and vice versa).

For any two mutually exclusive events, the probability that an outcome will be in one event or the other event is the sum of their individual probabilities

If A and B are mutually exclusive events, then according to the law of probability,

**⇒ P(A or B) = P(A U B) = P(A) + P(B)**

**Mutually Exclusive Events: Illustrations**

**iii) A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either 7 or 11?**

Sample space= rolling of two dies = 6 × 6= 36 Six outcomes have a sum of 7:(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) P(7) = 6/36

Two outcomes have a sum of 11: (5, 6), (6, 5) P(11) = 2/36 As you can observe, the sum of the numbers cannot be 7 and 11 at the same time, so these events are mutually exclusive. P (7 or 11) = P(7) + P(11) = 6/36 + 2/36 = 8/36 = 2/9

When 2 events are not mutually exclusive, there will be some overlapping between events as shown below. We now need to consider the overlapped region as well

For any two events, which are not mutually exclusive, the probability that an outcome will be in one event or the other event is the sum of their individual probabilities minus the probability of the outcome being in both events.

If events A and B are NOT mutually exclusive,

P(A or B) = P(A U B) = P(A) + P(B) – P(A ∩ B)

Where A ∩ B is the occurrence of A and B.

**iv) In the throw of a die, find the probability of getting an even number or a prime number The probability of getting an even number (2, 4, 6) = 3/6 = 1/2. The probability of getting a prime number (2, 3, 5)= 3/6 = 1/2**

Probability of getting a number which is both even and prime (2) = 1/6 P(even or prime) = P(even) + P(prime) – P(even and prime) = 1/2 + 1/2 – 1/6 = 5/6.

**Independent Events:**

Two events are said to be independent if the result of the second event is not affected by the result of the first event.

If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events. This is another law of probability which cn be defined as below:

If A and B are independent events,

#### P(A and B) = P(A) × P(B).

**Independent Events: Illustration**

**v) A drawer contains 3 red balls, 4 green balls, and 5 blue balls. One ball is taken from the drawer and then replaced. Another ball is taken from the drawer. What is the probability that the first ball is red and the second ball is blue?**

Because the first ball is replaced, the sample space of 12 balls does not change from the first event to the second event. The events are independent. P (red followed by blue) = P (red) × P(blue) = 3/12 × 5/12 = 15/144 = 5/48 .

**Dependent Events:**

If the result of one event is affected by the result of another event, the events are said to be dependent. If A and B are dependent events, the probability of both events occurring is the product of the probability of the first event and the probability of the second event once the first event has occurred. If A and B are dependent events, and A occurs first,

P(A and B) = P(A)× P(B, once A has occurred)

**Example:**

**vi) A drawer contains 3 red balls, 4 green balls, and 5 blue balls. One ball is taken from the drawer and is NOT replaced. Another ball is taken from the drawer. What is the probability that the first ball is red and the second ball is blue? **

Because the first ball is NOT replaced, the sample space of the second event is changed. The sample space of the first event is 12 balls, but the sample space of the second event is now 11 balls. The events are thus dependent. P (red then blue) = P (red) × P (blue) = 3/12×5/11 = 15/132=5/44.

**CAT Probability Practice Probability Questions:**

**1.** A pair of dice is rolled. Two possible events are rolling a number greater than 8 and rolling an even number. Are these two events mutually exclusive events?

**2.** A pair of dice is rolled. A possible event is rolling a multiple of 5. What is the probability of the complement of this event?

**3.** A paper bag contains 15 cubes of equal size. Eight of them are blue and are numbered from 1 to 8. Seven of them are pink and are numbered from 1 to 7. A cube is drawn from the bag and then replaced. A second cube is drawn. Are these two events independent?

**4.** A pair of dice is rolled. Two possible events are rolling a number which is a multiple of 3 and rolling a number which is a multiple of 5. Are these two events mutually exclusive?

**5.** A paper bag contains 15 cubes of equal size. Eight of them are blue and are numbered from 1 to 8. Seven of them are pink and are numbered from 1 to 7. What is the probability of drawing a cube with an even number?

**Theory Of Conditional Probability:**

This is based on the concept of Dependent events, where the probability that an event B takes place will depend on whether another event A has or has not taken place

The conditional probability of an event B, given the event A; denoted by P(B/A) is defined as

P(B/A) = P(A ∩ B)/P(A) where P(A) ≠ 0

P(A ∩ B) is the joint probability of A and B, or the probability of both A and B together.

Let’s understand this formula with an example:

**vii) A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? **

This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. P(Second/First) = P(First ∩ Second)/P(first) = 0.25/0.42 = 0.60 = 60%.

**viii) The probability that it is Thursday and that a student is absent is 0.03. There are 5 school days in a week. What is the probability that a student is absent given that today is Thursday? **

Since there are 5 school days in a week, the probability that it is Thursday is 0.2

P(Absent/Thursday) = P(Absent ∩ Thursday)/P(Thursday) = 0.03/0.2 = 0.15 = 15%.

**ix) At a CBSE school, 18% of all students play football and basketball and 32% of all students play football. What is the probability that a student plays basketball given that the student plays football?**

B → Basketball F → Football

P(B/F) = P(B ∩ F)/P(F) = 18/32 = 0.56 = 56%

**CAT Conditional Probability Practice Questions:**

**6.** In Del dorado, 84% of the houses have a garage and 65% of the houses have a garage and a back yard. What is the probability that a house has a backyard given that it has a garage?

**7.** 56% of all children get an allowance and 41% of all children get an allowance and do gardening. What is the probability that a child does gardening given that the child gets an allowance?

**8.** The probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68. What is the probability that a student takes Spanish given that the student is taking Technology?

At Byju’s the detailed CAT Syllabus and study materials are provided along with various video lectures to help students to prepare for CAT exam more effectively. To practice more of related topics, visit Probability Questions.