Probability is a chapter that is a simple and fun topic that uses lots of logic, this is tested in CAT. There are multiple ways to solve probability questions. You can get the answers using your way without even using formulae. However, using the formulae for probability will make your job easier.

**What is Probability?**

One cannot predict all the events with total certainty. The best we can say is, how likely are they to happen, using the idea of probability. Probability is the branch of mathematics that tells us the possible outcomes of given events divided by the total number of possible outcomes. Generally, the word “probability” is used to mean the chance that a particular event (or set of events) and has a value from 0 to 1, it can also be written as a percentage between 0 and 100%.

**Importance of Probability in CAT**

The concept 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 questions on the probability 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)

Different model problems from Permutations and Combinations from older CAT papers

1. A and B take part in a duel. A can strike with an accuracy of 0.6. B can strike with an accuracy of 0.8. A has the first shot, post which they strike alternately. What is the probability that A wins the duel?

A can win in the following scenarios:

A strikes in the first shot.

A misses in the first shot, B misses in the second, A strikes in the third.

A misses in the 1st shot, B misses in the 2nd, A misses the 3rd, B misses the 4th and A strikes the 5th.

And so on…

P(A in 1st shot) = 0.6

P(A in 3rd shot) = 0.4 * 0.2 * 0.6 {A misses, then B misses and then A strikes}

P(A in 5st shot) = 0.4 * 0.2 * 0.4 * 0.2 * 0.6 {A misses, then B misses and then A misses, then B misses, then A strikes}

Overall probability = Sum of all these

= 0.6 + 0.4 * 0.2 * 0.6 + 0.4 * 0.2 * 0.4 * 0.2 * 0.6 + …

Which is nothing but

0.6 + (0.4 * 0.2) * 0.6 + (0.4 * 0.2)2 * 0.6 + (0.4 * 0.2)3 * 0.6…

his is an infinite geometric progression with first term 0.6 and common ratio 0.4 * 0.2

Required Probability = a/1−r = 0.6/1−0.08 = 0.6/0.92 = 30/46 = 15/23

Hence the answer is “15/23”

2. Doctors have devised a test for leptospirosis that has the following property: For any person suffering from lepto, there is a 90% chance of the test returning positive. For a person not suffering from lepto, there is an 80% chance of the test returning negative. It is known that 10% of people who go for testing have lepto. If a person who gets tested gets a +ve result for lepto (as in, the test result says they have got lepto), what is the probability that they actually have lepto?

Let us draw the possibilities in this scenario.

Prob (patient having lepto) = 0.9

Prob (patient not having lepto) = 0.1

Given that patient has lepto, Prob (test being positive) = 0.9

Given that patient has lepto, (Prob test being negative) = 0.1

Given that patient does not have lepto, Prob (test being negative) = 0.8

Given that patient does not have lepto, Prob (test being positive) = 0.2

Now, we are told that the test turns positive. This could happen under two scenarios – the patient has lepto and the test turns positive and the patient does not have lepto and the test turns positive.

Probability of test turning positive = 0.9 * 0.1 + 0.9 * 0.2 = 0.27.

Now, we have not been asked for the probability of test turning positive. We are asked for the probability of patient having lepto given that he/she tests positive. So, the patient has already tested positive. So, this 0.27 includes the set of universal outcomes. Or, this 0.27 sits in the denominator.

Within this 0.27, which subset was the scenario that the patient does indeed have lepto?

This is the key question. This probability is 0.1 * 0.9 = 0.09.

So, the required probability = 0.09/0.27 = 1/3

So, if a patient tests positive, there is a 1 in 3 chance of him/her having lepto. This is the key reason that we need to be careful with medical test results.

Hence the answer is “1/3”

3. If all the rearrangements of the word AMAZON are considered, what is the probability that M will feature between the 2As?

A. For this type of question, we need to consider only the internal arrangement within the M and 2As.

M and 2As can be rearranged as AMA, AAM, or MAA.

So, the probability that M will feature between the 2As is 1/3.

Now, let us think why we need to consider only the M and 2As.

Let us start by considering a set of words where the M and 2 As are placed at positions 2, 3 and 5.

The other three letters have to be in slots 1, 4 and 6

Three letters can be placed in three different slots in 3! = 6 ways.

Now with __ M A __ A __ there are 6 different words.

With __ A M __ A __ there are 6 different words.

With __ A A __ M __ there are 6 different words.

For each selection of the positions for A,A and M, exactly one-third of words will have M between the two A’s.

This is why only the internal arrangement between A, A and M matters.

So, probability of M being between 2 As is 1/3.

Hence the answer is “1/3”

4. N is a 3-digit number that is a multiple of 7; what is the probability that it will be a multiple of 5?

A. N is a three digit multiple of 7.

N could be 105, 112, 119, 126…..994.

Or, 15 * 7, 16 * 7…..142 * 7.

Or there are 142–14 = 128 numbers.

Within these we need to locate the multiples of 5.

Or, we need to isolate multiples of 35.

Or, we need to see how many numbers there are in the list 105, 140, 175…..980.

35 * 3, 35 * 4…35 * 28…

Or, there are 26 such numbers.

Probability 26/128 = 13/64

5. A boss decides to distribute Rs. 2000 between 2 employees. He knows X deserves more that Y, but does not know how much more. So he decides to arbitrarily break Rs. 2000 into two parts and give X the bigger part. What is the chance that X gets twice as much as Y or more?

The bigger part could be any number from 1000 to 2000.

Now, if the bigger part is to be at least twice as much as the smaller part, we have

X ≥ 2Y or X ≥ 2(2000 – X)

Or X ≥ 4000/3

Given that X lies between 1000 and 2000, what is the probability that X lies between 4000/3 and 2000?

This probability is equal to (2000−4000/3)/(2000−1000) = 2/3

Hence the answer is “23”

6. A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag one ball is drawn, then what is the probability that the ball drawn is red?

Let Red balls be ‘r’ and brown balls be ‘b’

P(r) = p(b1) * p(r) + p(b2) * p(r)

P(R) = 1/2 * 4/7 + 1/2 * 2/5

P(R) = 17/35

Hence the answer is “17/35”

**Level-1 preparation**

- 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
- 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 questions and try to differentiate the type of event present in the question to get a clear cut understanding before moving forward

**Level-2 preparation**

- Practice different types and model questions to gain confidence.
- 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

**Easy-difficulty level questions**

- 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

**Moderate-difficulty level 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

**Hard-difficulty level questions**

- 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

**Level-3 preparation**

- 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 which are on a hard level to get a gist of the type of problem that can be asked
- Go through advanced level material 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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