# Probability For CAT Concepts And Practice Questions

## Learn all the theorems of probability for CAT and solve practice questions in this article. By IIM Skills – online CAT coaching.

Probability for CAT is an important topic in the CAT quantitative aptitude section. You can expect 2-3 questions from probability in the quant section in CAT. Probability helps us determine the odds of an event happening or not happening.

“Probability is the logic of uncertainty and randomness. Uncertainty and randomness happen in just about every field of application and in daily life, so it is greatly useful and also fascinating to understand probability.”

As a manager, it is an important topic that helps us learn the possibilities of success of a business decision for an organization. It helps us make better decisions by weighing the pros and cons of a judgment. It is always our goals with any decision, that the result of the decision we make today will be in our favor.

Many aspects of business, finance, and society posses a level of uncertainty on the outcomes that may or may not be the same in every set of conditions. The outcomes lack deterministic regularity. In our regular lives and day to day activities, we use the principles of probability, for instance, will it rain today, will the price of that particular share go up, which team will win the match, what are the chances that the sales will go up by 12 percent this month, and so on. These are small ways where we attempt to determine the probability of an event occurring.

All these questions aim to find the chance or simply, the probability or the likelihood of an event happening. Probability and its principles give us tools, mathematical formulae, and statistics, to determine the odds of occurrence of an event

The probability of an event is a number between 0 and 1. 0 means the event has absolutely no chance of occurring. While 1 means that the event is certain to happen. The higher the probability of an event, more is the likelihood of it occurring.

Probability is denoted by P(E) where P is the probability of event E happening. It’s indisputable that probability is a very helpful and valuable mathematical and statistical resource. The uses of principles of probability are everywhere. It is wanted in every area whether physics, commerce, finance etc.

Since, this concept is so fundamental and important, as future managers, you need to properly study the principles of probability for CAT and understand it’s applications. For this reason, it has been included and frequently asked every year in the Quantitative Aptitude section of CAT.

Probability for CAT is a simple and fun topic that uses lots of logic. There is no one correct way to solve probability questions. You can arrive at the right answer in your way, even if you choose not to use any probability formulae for CAT. However, the formulae for probability only make your job easier and it is advisable by IIM Skills- online CAT coaching that you learn and memorize them when you are preparing for CAT.

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Before we dive in to practice question for probability for CAT, let’s learn the theory of probability for CAT quant section.

Below are some common terms and definitions that we will use in the article and you will also encounter them elsewhere during your CAT preparation.

Event: Event refers to any outcome of an experiment that is independent of any other outcome and cannot occur simultaneously with another event.

Experiment: An action that can deliver some well-defined outcomes, is an experiment.

Random Experiment: “ If every trial of an experiment is carried out under identical conditions, and the outcome is not unique but maybe one of many possible outcomes then such experiment is known as a random experiment.”

Sample Space: The set of all probable outcomes in a Random Experiment is referred to as Sample space S, on the condition that no two of these outcomes can occur together and precisely one of these results must happen whenever the experiment is carried out.

Certain and Impossible events: If S is a sample space, both S and the null set φ are events. S is termed as a certain event and φ is termed an impossible event.

Equally Likely Events: The given set of events is equally likely only when the occurrence of one event is not more probable than any other event.

Exhaustive events: In probability theory, a set of events is said to be exhaustive when at least one of the events in that set occurs every time the experiment is carried out.  There is no other possible event outside the set of events. For example, flipping a coin has only two outcomes ie. head or tails. There is no other possibility.

Mutually Exclusive events: A set of events is said to be mutually exclusive when the occurrence of one event automatically means the impossibility of another event happening on that experiment. Two events A and B are said to be mutually exclusive if AꓵB = φ.

Algebra of Events: Let A and B are two events. Then,

1. The event ‘either A or B’ happens if at least one of A and B occurs. It is usually denoted as AꓴB (A or B).
2. The event ‘both A and B’ happens if both A as well B happen simultaneously. It is depicted as AꓵB (A and B).
3. The event ‘A not’ is said to happen if A does not happen. It is denoted as Aᶜ.
4. The event ‘A but not B’ happens if A occurs but B does not happen. It is indicated by A-B.

These were the common terminology that we are going to use in this article about probability for CAT. Now, let us look at some formulae and concepts of probability.

Probability of an event: It essentially means during the execution of a random experiment the occurrence of any event is always random but a measure of its likely occurrence can be calculated and that is known as the probability of the event.

1. The probability = 0; for a null event
2. The probability= 1 for a sure event
3. 0 ≤ P(E)≤1 by definition
4. ∑P(E)= 1  since a set of events is exhaustive

No. of favorable outcomes for event A

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

For example. If a die is rolled, what is the probability of rolling a 4?

“The sample space of the above random experiment S is {1, 2, 3, 4, 5, 6} = 6 outcomes.”

Therefore, the probability of rolling a 4 is = 1 / 6.

Given here are some relevant principles and rules that will help you in understand probability problems easily.

• “If A is a subset of B then, P(A)≤P(B).”
• “P(φ) = 0.”
• “P(S) = 1.”
• “P(Aᶜ) = 1 – P(A).”
• “P(B-A) = P[B-(AꓵB)] = P(B)-P(AꓵB).”
• “P(AꓴB) = P(A) + P(B) – P(AꓵB).”
• “P(AꓴB) = P(A) + P(B) when P(AꓵB) = φ.”
• “P(AꓴBꓴC) = P(A) + P(B) + P(C) – P(AꓵB) – P(BꓵC) – P(CꓵA) + P(AꓵBꓵC).”

#### Conditional Probability:

Let S be a sample space. Let A and B be any two events. A ≠ φ.

Then, the probability of the occurrence of event B, if A has already occurred, is called “conditional probability of B restricted to the occurrence of A”. It is denoted as P(A/B).

“Thus, the probability of the event B restricted to the occurrence of the event A is the same as the probability of event AꓵB while A is considered as sample space.”

“P(B/A) = n(AꓵB)/ n(A) = P(AꓵB)/P(A)”

1. “P(AꓵB) = P(A). P(B/A)”

If A ≠ φ & B ≠ φ then,

2. “P(AꓵB) = P(A). P(B/A) = P(A). P(A/B).”

Conditional probability theory is a very significant theory of probability. It is used in many kinds of problems. Study the example below to understand the concept of conditional probability thoroughly.

Example: A card is drawn randomly from a pack of cards. Find the probability of the card being a heart if it is already a king?

Solution: Let A be the event where the card is of heart

and B be the event where the card is king

Consequently, P(A) = 13/52; and P(B) = 4/52

=> P(AꓵB) = 1/52

Thus, P(A/B) = P (King of heart) = P(AꓵB)/P(A)

= (1/52)/ (4/52)

= ¼

This was an easy example to clear the usage of conditional probability. Ideally, you should practice several questions to get a grasp and score high in probability for CAT.

Independent Events: Two events are assumed to be independent if the likelihood of occurrence of one event does not affect the likelihood of occurrence of the other i.e. two events A and B are said to be independent if

“P(A/B) = P(A/Bᶜ) = P(A)”

“Or, P(B/A) = P(B/Aᶜ) = P(B)”

“Or, P(AꓵB) = P(A). P(B)”

Let us consider another example:

Example: A puzzle is given to three students. The chances of solving for each student are ½, ⅓, and ¼ respectively. Find the probability that the puzzle is going to be solved?

Solution:

Let there be 3 events A, B, and C such that

A is the event where the student with probability ½ solves the puzzle,

B is the event where the student with probability ⅓ solves the puzzle,

C is the event where the student with probability ¼ solves the puzzle,

So, P(A) = ½,

P(B) = ⅓, and

P(C) = ¼

Accordingly, Aᶜ, Bᶜ and Cᶜ are the events when the respective students could not solve the puzzles.

P(Aᶜ) = 1-½ = ½,

P(Bᶜ) = 1-⅓ = ⅔,

P(Cᶜ) = 1-¼ = ¾

Since we are required to determine the probability of solving the puzzle and to find out the probability of not solving the puzzle, we just subtract the former from 1 to get the answer.

As we can observe, in this example, the probabilities of every event is independent of other events. More than one student can also solve the puzzle

Therefore, P(AᶜꓵBᶜꓵCᶜ) = P(Aᶜ). P(Bᶜ). P(Cᶜ) = ½ X ⅔ X ¾ = ¼

Hence, the P [succesful solution of the puzzle] = 1 – P [no student solves the puzzle]

=> 1 – ¼ = ¾

Note:  Three events A, B, and C are said to be independent of each other if

“P(AꓵBꓵC) = P(A). P(B). P(C)”

Similarly,

Three events A, B, and C are said to be pairwise independent if

“P(AꓵB) = P(A). P(B), P(BꓵC) = P(B). P(C) & P(AꓵC) = P(A). P(C)”

#### The connection among Independence and Mutually Exclusiveness of any two events.

• “If two events A ≠ φ and B ≠ φ are independent, then they are not mutually exclusive.”
• “If two events A ≠ φ and B ≠ φ are mutually exclusive, then they are not independent.”

Independent Experiments:  Let there be two random experiments where one follows the other. If on repeated conducting of experiments, the sample space of any one event is not influenced by the outcome of the other, then two experiments are independent of each other.

It is important to note that there’s a distinction between independent events and independent experiments. Independent events define events within one experiment whereas independent experiments are two or more separate experiments with their own set of events.

i.e. consider the rolling of a dice, here the sample space for a second or third roll is not influenced by the results of the first roll. So, two rolls are random experiments.

Bayes’ Theorem:

This is an important theorem and it is possible that it may be used in questions in probability for CAT.

Suppose A₁, A₂, … An, are n mutually exclusive and exhaustive events in a given set. Thus, they divide the sample of events into n parts and event B occurs.

Then the conditional probability that Ai occurs provided B has already happened is:

“P(A/B) = P(Ai ). P(B/Ai )”

“∑ni=1 P(Ai ). P(B/Ai )”

Binomial distribution of successive events

This is a concept of statistics that is sometimes used in solving questions in probability to simplify the calculations.

Let us assume p and q are the probabilities of the occurring and not-occurring of an event on a single trial.

Then the chance of it occurring r times in n trials is nCrprqn   – r because the chance of its occurring r times and not-occurring n – r times in given order is prqn r and there are nCr such orders which are mutually exclusive and for any such order, the probability is prqn  –r

Therefore, the required probability is nCrprqn  – r. Now let’s learn about its use with an example.

Example: A fair die is rolled until six comes up three times. The probability that exactly 5 rolls are required will be?

Now the probability of rolling a ‘six’ on the die in the first trial is 1/6. Therefore ‘p’ is 1/6 and thus, q is 5/6.

=>So we want six three times in precisely 5 trials. So, let’s set that in 5th trial six shows up and its probability will be 1/6.

=>Consequently, the probability of getting six two times out of the 4 trials using the above distribution will be = 4C2 *(5/6)2 * (1/6)2.

Consequently, the probability of getting three ‘six’ in precisely 5 tosses is [ 4C2 *(5/6)2 * (1/6)2] * 1/6 = 25/1296.

CAT Practice questions for probability

For questions in probability, since many times there are no specific ways of solving, you will get better only if you practice more.

Question 1: Sum of three numbers a, b and c is 10. How many ordered triplets (a, b, c) can exist provided a,b,c are natural numbers?

1. 45
2. 36
3. 54
4. 28

Solution: Here,

a + b + c = 10. Now, let us draw ten lines

|       | |       | | |       | | | |

This question now essentially becomes about placing two ‘+’ signs between these sticks. For instance,

|       | |       | + | |       | | | + |

This would be the equal to 4 + 5 + 1. or, a = 4, b = 5, c = 1.

There are 9 openings between the lines, out of which we have to select 2 for inserting the ‘+’s.

The number of ways of performing this would be 9C2. This calculation counts ordered triplets. (4, 5, 1) and (1, 4, 5) are both calculated as separate possibilities.

We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural numbers as in this question) with a small change to the above approach. Give it some thought.

Hence the correct answer is (B) 9C2 = 36.

Question 2: In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?

1. 72
2. 54
3. 45
4. 36

Solution:

This question only requires you to calculate the number of ways of a, b, c such that a + b + c = 11, where a, b, c are natural numbers. By making them natural numbers, we see that no number can be zero

10C2 = 45 ways

This is similar to the previous question.

Hence the correct answer is (C) “45”

Question 3: “If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?”

1. 4716
2. 4720
3. 4718
4. 1717

Solution:

We don’t have to worry about finding meaningful words.

The first word would be ABCDE. With 2 different vowels, 3 different consonants, this is the first word that can be made.

Starting with AB, we can make any number of words.

AB __ __ __.

The next three spaces must have exactly 2 consonants and 1 vowel. This can be chosen in 20C2 and 4C1 ways. Then the three different letters can be rearranged in 3! = 6 ways.

Or, number of arrangements beginning with AB = 20C2 * 4C1 * 3! = 190 * 4 * 6 = 4560

Next, we consider words beginning with ACB

ACB __ __.

The last two spaces have to be packed with one vowel and one consonant. = 19C1 * 4C1. This order can be rearranged in 2! = 2 different ways.

Or, number of words starting with ACB = 19C1 * 4C1 * 2 = 19 * 4 * 2 = 152

Next, we move on to words beginning with ACDB: There are 4 different arrangements on this type– “ACDBE, ACDBI, ACDBO, ACDBU”

So far, the number of words we arranged = 4560 + 152 + 4 = 4716

Starting with AB    => 4560

Starting with ACB    => 152

Starting with ACDB    => 4

Total words      => 4716

After this we move to words starting with ACDE, the first possible word is ACDEB. After this we have ACDEF.

So, the rank of ACDEF alphabetically = 4718

Hence the correct answer is (C) “4718”

Question 4: “From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?”

1. 36
2. 60
3. 84
4. 72

Solution:

We can agree that any number that is a multiple of 2 and 6 is also a multiple of 12?

Any number that is a multiple of 12 should also be a multiple of 4 and 3.

First, let us look at the condition for a number to be a multiple of 3.

Sum of the numbers should be a multiple of 3.

Here, the sum of every number from 2 to 7 is 27.

So, if we need to omit one digit and still be a multiple of 3, we need to omit either 3 or 6.

So, the likely 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.

When the digits are 2, 4, 5, 6, 7.

The last two digits for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76.

For each of these combinations, there are 6 different arrangements possible.

So, with this set of 5 digits we can have 36 different numbers.

When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24.

For each of these combinations, there are 6 different numbers possible.

So, with this set of 5 digits, we can have 24 different arrangements.

Overall, there are 60 different 5-digit numbers arrangements possible.

Hence the correct answer is (B) “60”

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###### Author: Gaurav Rayakwar
Gaurav is a Content Writer at IIM Skills. He has a B.Tech. degree but then he switched to the creative side by doing his master's in advertising and public relations. Gaurav is also a part-time blogger and graphic designer currently living in Mumbai