- Sequential Output Tracing (Input Output)
- Syllogism
- Critical Reasoning: Assumptions & Conclusions
- Relationships
- Cause & Effect
- Circular Arrangement
- Coding-Decoding
- Complex Arrangements
- Critical Reasoning: Course of Action
- Critical Path
- Direction and Word Puzzle
- Games And Tournaments
- Linear Arrangement
- Team Selection
- Venn Diagram
- Bar Graphs
- Data Sufficiency
- Line Graph
- Pie Chart
- Tables and Caselets

The concept of probability is so fundamental and important. As future managers, you need to properly study the principles of probability and understand its applications. The quantitative section of the CAT contains probability aptitude questions to test you on this.

• There were no questions asked directly on probability in the past 3 years. But this is a concept that can often be useful to solve some other concepts like Permutations and Combinations, Venn Diagrams, etc.

**Importance of Probability in XAT**

Year | No. of Questions | Difficulty |

2020 | 1 | Hard |

2019 | 0 | – |

2018 | 1 | Moderate |

2017 | 1 | Easy |

**Importance of Probability in SNAP**

Year | No. of Questions | Difficulty |

2018 | 0 | – |

2015 | 2 | Moderate |

2014 | 4 | Easy-Moderate |

2013 | 1 | Moderate |

__Importance of Probability in IIFT__

Year | No. of Questions | Difficulty |

2019 | 1 | Moderate |

2018 | 1 | Hard |

2017 | 1 | Moderate |

2016 | 0 | – |

**Experiment:**It is an operation that can produce some well-defined outcomes**Random Experiment:**If each of the trials of an experiment is conducted under identical conditions, then the outcome is not unique, but maybe of any possible outcome then such an experiment is known as a random experiment.**Sample Space:**The set of all possible outcomes in a Random Experiment is known as Sample space, provided no two or more of these outcomes can occur simultaneously and exactly one of these outcomes must occur whenever the experiment is conducted.**Event:**Any subset of a sample space is called an event.**Certain and Impossible events:**If S is a sample space, then both S and null set φ both are events. S is called a certain event and φ is called an impossible event.**Equally Likely Events:**The given events are said to be equally likely if none of them is expected to occur in preference of the other.**Exhaustive events:**In probability theory, s system of events is called exhaustive, if at least one of the event of the system occurs. Ex. If a coin is tossed then Head and Tails form an exhaustive set of events.**Mutually Exclusive events:**A set of events is called mutually exclusive events if the occurrence of one of them precludes the occurrence of any of the remaining events. In other words, if there are two events P and B then they are called mutually exclusive if PꓵB = φ.**Independent events:**P and Q are two events. Then,

- The event ‘either P or Q’ is said to occur if at least one of P and Q occurs. It is usually denoted as P ꓴ Q (P or Q).
- The event ‘both P and Q’ is said to occur if both P as well Q occur simultaneously. It is represented as P ꓵ Q (P and Q).
- The event ‘P not’ occurs if P does not occur. It is written as Pᶜ.
- The event ‘P but not Q’ is said to occur if P occurs but Q does not occur. It is denoted by P – Q.

**Probability of an event:**It means in the performance of a random experiment the occurrence of any event is always uncertain but a measure of its probable occurrence can be devised known as the probability of the event.

- The probability of the null event is 0.
- The probability of a sure event is 1.
- 0 ≤ P(E)≤1
- ∑P(E)= 1

- Let S denotes the sample space of a random experiment and P be an event. Then the probability of P is defined as No. of favorable outcomes for event P

Total no. of outcomes. =n(P)/ n(S).

- Some important theorems

- If P is a subset of Q then, P(P)≤P(Q).
- P(φ) = 0.
- P(S) = 1.
- P(Pᶜ) = 1 – P(P).
- P(Q-P) = P[Q-(PꓵQ)] = P(Q)-P(PꓵQ).
- P(PꓴQ) = P(P) + P(Q) – P(PꓵQ).
- P(PꓴQ) = P(P) + P(Q) when P(PꓵQ) = φ.
- P(PꓴQꓴR) = P(P) + P(Q) + P(R) – P(PꓵQ) – P(QꓵR) – P(RꓵP) + P(PꓵQꓵR).

- Conditional Probability:

Let S be the sample space. Let P and Q be any two events. P ≠ φ. Then, the probability of event Q, if P has already occurred, is called the conditional probability of Q restricted to the occurrence of P. It is represented as P(P/Q). Thus, the probability of the event Q restricted to the occurrence of event P is the same as the probability of event PꓵQ while P is considered as sample space.

P(Q/P) = n(PꓵQ)/ n(P) = P(PꓵQ)/P(P)

P(PꓵQ) = P(P). P(Q/P)

If P ≠ φ & Q ≠ φ then,

P(PꓵQ) = P(P). P(Q/P) = P(P). P(P/Q).

- Conditional probability is a very important concept of probability. It is used in a variety of questions. Let’s ingrain the concept more comprehensively using the example given below.

Independent Events: Two events are said to be independent if the probable occurrence or non-occurrence of anyone is not affected by the occurrence or non-occurrence of the other i.e. two events P and Q are independent if

- P(P/Q) = P(P/Qᶜ) = P(P)
- Or, P(Q/P) = P(Q/Pᶜ) = P(Q)
- Or, P(PꓵQ) = P(P). P(Q)

The relation between Independent and Mutually Exclusiveness of two events. - If two events P ≠ φ and Q ≠ φ are independent, then they are not mutually exclusive.
- If two events P ≠ φ and Q ≠ φ are mutually exclusive, then they are not independent.
**Independent Experiments:**Let there be two random experiments one after the other. If on repeated performances the sample space of any is not affected by the result of the other, then two experiments are independent of each other.- There’s a difference between independent events and independent experiments, but students tend to confuse between the two. The former talks about events in an experiment and the latter explain independence among multiple experiments.
- Bayes’ Theorem: Suppose P₁, P₂, … Pn, are a mutually exclusive and exhaustive set of events. Thus, they divide the sample into n parts and event Q occurs. Then the conditional probability that Pi happen given that Q has happened is given by

P(P/Q) = P(Pi). P(Q/Pi)

∑ni=1 P(Pi). P(Q/Pi)

1. Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18,choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is (CAT 2020)

A) 19 B) 22 C) 20 D) 21

Let a, b, c represent number of students who opted for two subjects – Maths and Chemistry, Maths and Physics and Physics and Chemistry

respectively.

Given, a+b=23−18=5a+b=23−18=5

and b+c=25−18=7b+c=25−18=7

Since a, b and c cannot be negative the least value for any of the three is 5 .

We get a+c+18=(23+25−18)−2ba+c+18=(23+25−18)−2b

Minimum value of number of students who chose chemistry =23+25−18−10=20

2. A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is (CAT 2019)

A) 32 B) 38 C) 43 D) 45

2. As per question a+b+c + (d+e+f) + g =256—————-1)

From figure , a+b+c + 2(d+e+f) + 3g = 132+144+123 = 399 ———-2)

Also g+e = 58 ———x)

f+g = 25—————y)

and d+g = 63————z)

adding all three , d+e+f + 3g = 58+25 + 63 = 146 ————-3)

from eq 2) and 3) , a+b+c + d+e+f = 399 – 146

a+b+c + d+e+f = 253————4)

from eq 1 and eq 4) g = 256 – 253 =3

so from eq x) , y) and z) d= 60, e = 55 and f = 22

Number of people playing tennis = 123

So g + f + e + c = 123

3+22 + 55 + c = 123

Or c = 43

the number of members who can play only tennis = 43

3. A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible? (CAT 2006)

(1) 3 (2) 4 (3) 5 (4) 6 (5) 7

Let the no. of students in front row be x.

So, the no. of students in next rows be x – 3, x – 6, x – 9…. so on

If n i.e. no. of rows be 3, then x + (x – 3) + (x – 6) = 630

3x = 639

x = 213 So possible.

Similarly, for n = 4,

x + (x – 3) + (x – 6) + (x – 9) = 630

4x – 18 = 630

X= 648/4 = 162

x = 4 to possible. If n = 5,

(4x – 18) + (x – 12) = 630

5x – 30 = 630

x = 120

Again n = 5 is possible. If n = 6,

(5x – 30) + (x – 15) = 630

6x – 45 = 630

6x = 675

x ≠ Integer Hence, n ≠ 6.

4. At a bookstore, ‘MODERN BOOK STORE’ is flashed using neon lights. The words are individually flashed at the intervals of 2(1/2)s, 4(1/4)s and 5(1/8)s respectively, and each word is put off after a second. The least time after which the full name of the bookstore can be read again is (CAT 2002)

1. 49.5 s 2. 73.5 s 3. 1744.5 s 4. 855 s

• Initially try solving the small questions without using any concept or formula. It’ll help you understand why and how the concept is being derived to solve them You can refer MBAP CAT E Book (Concept Theory) study material and MBAP Live Lecture Recording (Basic) on basic concept for more clarity.

• Learn the basic concepts using the cheat sheets available everywhere. You can refer to the concepts mentioned above in this article

• Learn the types of events and what they mean.

• Try solving basic and CAT questions using MBAP CAT E Book (Practise Questions) and MBAP lecture Assignment, and try to differentiate the type of event present in the question to get a clear cut understanding before moving forward

• Practice different types and model questions from MBAP CAT E Book (Practise Questions) and MBAP lecture Assignment to gain confidence using.

• Go through multiple solved questions to get different approaches to a question

• Once you get the answer to a question, try plugging in the value to verify your answer

• Questions are pretty straightforward. There won’t be any complex statements. Usually a single line question

• Basic concepts learned will help to solve these very easily

• Candidates should concentrate on types of events, and some knowledge on permutations and combinations

• Knowing the concepts covered above would be handy as you will be able to solve the question very quickly and save time for other questions

• Questions might be tricky to look at. Once you understand the ask in the question and able to map it to the correct concept it’ll be easy to solve it

• Concept and formula knowledge is very important as it helps you to complete the problem smoothly without any barriers

• Keep practicing multiple problems to get a vast idea of all the different types of questions. This will help you to map the formula to the question of understanding the question easy

• Unlike easy questions, you may have to apply multiple concepts in a single question

• For getting admission in the dream IIMs, you need to have a percentile of 99%. These level-3 questions differentiate candidates between a 90 percentile and a 99 percentile

• Questions might be the length, but the key is to keep yourself concentrated and understand all the facts or numbers given in the question which will be useful to solve the questions

• Multiple concepts may be required to solve a question, identifying the right formula or concept and the order in which you must apply is the key

• Practice multi-topic questions like permutations and combinations mixed with probability, you can often find these types of questions in competitive exams

• Keep solving previous year CAT questions using MBAP Previous year CAT paper which are on a hard level to get a gist of the type of problem that can be asked

• Go through advanced level material of MBAP CAT Advance E books to understand high-level concepts

• Try solving good books like Arun Sharma to have an idea on types of questions that can be asked

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