CAT Probability Questions – Concepts, Formulas, Expert’s Tips to Prepare | Download Probability CAT Questions PDF

What are Probability Questions for CAT?

Considering the trends, Probability CAT Questions are a significant topic for the QA section. The candidates can expect 2-3 questions, and the difficulty level will be from moderate to high. So, it is important to know the concepts and formulas of Probability CAT Questions to answer these questions accurately.

Probability Questions for CAT are quite simple if the candidates practice well. It only uses logic, although you have chosen the correct formula of Probability CAT Questions. Note the table below section to get a detailed overview of Probability Questions for CAT.

☛ Candidates must check all the details related to CAT Exam if they are thinking of joining IIM to study MBA.

Types of Probability CAT Questions

To solve CAT Probability Questions, one must have a clear understanding of the concepts of Permutation and Combination CAT Questions. To master this area, one must brush up on their knowledge of permutation and combination. Probability CAT Questions are logical in nature, and one should be aware of the types of questions to solve them easily. Below mentioned are some of the types of Probability CAT Questions:

1. Bayes Theorem
2. Dependent Events/ Independent Events/ Mutually Exclusive Events
3. Conditional Probability
4. Multiplication Rule of Probability

Formulae for CAT Probability Questions

The probability formulae is used to compute the probability of an event. In other words, the likelihood of an event happening is called probability.

1. Random Experiments: An experiment is called a random experiment if it satisfies the following two conditions: (i) It has more than one possible outcome. (ii) It is not possible to predict the outcome in advance.
2. Outcomes: A possible result of a random experiment is called an outcome.
3. Sample Space: The set of all possible outcomes is called Sample Space.
4. Probability of an event A [P(A)] = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 favourable? 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐓𝐨𝐭𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 Or, P(A) = 𝐧(𝐀) 𝐧(𝐒) Where ➢ P(A) is the probability of an event “A”. ➢ n(A) is the number of favourable outcomes of event A. ➢ n(S) is the sample space Ex: What is the probability of getting a jack from a well-shuffled pack of cards? Sol: Here, n(S) = 52 and n(E) = 4 (Total 4 jacks) Hence, required probability = P (E) = n(E) n(S) = 4 52 = 1 13 .
5. Range of probability of an event A: 0 ≤ P(A) ≤ 1
6. Odds: a. Odds in favour = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 favourable? 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 unfavourable? 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 b. Odds against = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 unfavourable? 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 favourable? 𝐨𝐮𝐭𝐜𝐨𝐦𝐞?
7. Addition Rule: P(A U B) = P(A) + P(B) – P(A∩B)
8. Multiplication Rule: P(A ∩ B) = P(A) P( B A ) = P(B) P( A B )
9. Equally likely events: When two or more events have an equal chance of happening or equal probability of occurrence, then those events are called equally likely events.
10. Independent events: When two events have occurred, then the probability of occurrence of one event is not affected by the occurrence and non-occurrence of the other event, and then the two events will be called the independent event. For Independent events A and B: P( B A ) = P(B) and P( B A ) = P(B) So, P(A ∩ B) = P(A). P(B)
11. Complementary events: When two events have occurred, such that if one of the events happens, then the other event cannot happen and vice versa. The complimentary event of event A is denoted by A’ or A̅. For complimentary events A and A’ or A̅: P(A) + P(A’) = 1
12. Exhaustive events: When two or more events have occurred. then all such that if one of the events happens, then the other event cannot happen and vice versa. For Exhaustive events A, B, C, … : P(A) + P(B) +P(C)… = 1
13. Mutually exclusive events: Two events are mutually exclusive if they cannot occur at the same time. In other words, mutually exclusive events are disjoint events. So, the intersection of two disjoint events is an empty event. If two events are disjoint, then the probability of them both occurring at the same time is 0. If events E1 and E2 are mutually exclusive events, then
14. a. E1 E2 = b. P(E1 E2) = 0 c. P (E1 U E2) = P(E1) + P(E2)
15. Conditional Probability: P(A | B) = P(A ∩ B) P(B)

Example: Calculate the probability of getting an even number if a dice is rolled.

Answer: ½

Solution: Sample space (S) = {1,2,3,4,5,6}, so, n(S) = 6

Let E be the event of getting an even number, then

E = {2,4,6} so, n(E) = 3

Required probability = n(E)/n(S) = 3/6 = ½

Probability Questions for CAT PDF

It is advised to the candidates to practice Probability CAT Questions in order to solve questions on Modern Mathematics. Solving past years’ papers and CAT Mock Tests will help the candidates to understand the different types of questions that can be asked on this topic. Candidates can download the Probability CAT Questions PDF given below:

☛ Probability Questions for CAT PDF Download

Probability CAT Questions Sample

Listed below are some of the sample CAT Questions for Probability. The candidates can go through the below-mentioned questions to ace the QA for CAT. The Probability CAT Questions given below will provide the candidates with an overview of the type of questions that are asked on this topic.

You must also go through CAT Probability Questions from the previous year to gain a better understanding of the kind of questions asked. Download the CAT Question Paper to practice Probability CAT Questions again and again.

Question 1: In the MBA Programme of a B – School, there are two sections, A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as a class representative, the probability that exactly one of the students chosen is a girl is:

Answer: 17/36

Explanation: Selecting a girl from section A and section B is 1/4 and 4/9, respectively.

Selecting a boy from section A and section B is 3/4 and 5/9, respectively.

Case 1: A girl from section A and a boy from section B.

P1 = (¼) X (5/9) = 5/36

Case 2: A boy from section A and a girl from section B.

P2 = (¾) X (4/9) = 12/36

Required Probability = P1+P2 = 17/36

Question 2: If 2 dice are thrown simultaneously, what is the probability that 1 of them shows up ‘2’ and the other shows ‘5’?

Answer: 1/18

Explanation: Total number of outcomes = 6 x 6 = 36

There are 2 favourable circumstances = (2,5) and (5,2)

Hence, 2/36 = 1/18

Question 3: A box contains 5 green, 3 black, and 7 red balls. Two balls are selected at random without replacement from the box. What is the probability that both balls are red?

Answer: 1/5

Explanation: Total number of balls = 5 + 3 + 7 = 15.

The number of ways in which 2 balls can be selected = 15C2 = (15×14)/(2×1) = 105

Total number of red balls = 7

The number of ways in which 2 red balls can be taken out = 7C2 = (7×6)/(2×1) = 21

The probability of getting 2 red balls = 21/105 = 1/5.

Alternative solution:

The probability of the first ball being red = 7/15.

If the first ball taken out is red, the box now has a total of 14 balls, of which 6 are red.

The probability of the 2nd ball being red = 6/14

The probability of both balls being red is (7/15)×(6/14) = 1/5.

Note: Here, we have to get a red ball in each of the draws, and hence we do not have to consider any arrangement, but if we draw balls of different colours, the arrangement will have to be taken into account.

Question 4: If P(A|B) = 1/3, P(B) = 1/4, and P(A) = 1/2, the probability that exactly one of these events, A and B, occur is:

Answer: 7/12

Explanation:

Or

So, we have P(A) = 1/2 and

Therefore, P (only A) = P(A) – P(A ⋂ B)

= 1/2 – 1/12 = 5/12.

Similarly, P(B) = 1/4 and P(A ⋂ B) = 1/12

Therefore, P(only B) = P(B) – P(A ⋂ B)

= 1/4 – 1/12 = 1/6.

Probability that exactly one of A and B happens = P(only A) + P (only B)

= 5/12 + 1/6 = 7/12.

Question 5: What is the probability of rolling three six-sided dice and getting a different number on each die?

Answer: 5/9

Explanation: For the first die, we can roll any one of six numbers.

For the second die, we can roll any number save for the number we rolled on the first die, giving us 5 possibilities.

For the third die, we can roll four different numbers (we can’t roll the number we rolled on the first or second die.

6 x 5 x 4 = 120 possibilities out of 216 total possibilities. (For total possibilities, we get 6 x 6 x 6 = 216).

120/216 = 5/9

Best Books to Study Probability CAT Questions

In order to ace the Quantitative Ability, the candidates need to refer to the best CAT Books and study material. Candidates can also refer to the NCERT books of classes 9 and 10 to practice Probability CAT Questions. The candidates must refer to the books which contain all the types of questions related to CAT Probability Questions.

Tips to Solve Probability CAT Questions

The candidates aspiring to take the CAT, need to know the tips and tricks to solve CAT Probability Questions. Candidates should go through the below-mentioned points in order to know the tips to solve Probability CAT Questions.

• Candidates should first start practising questions based on Permutation and Combination in order to solve Probability Questions for CAT quickly and easily.
• CAT consists of negative markings, so it is very important to choose the correct set of questions. As not many questions are asked from Probability, the candidates should only select those questions that they are sure about.
• The candidates should practice previous year’s papers and mock tests to know the types of Probability CAT Questions that can be asked on this topic.