*Saturday, October 7th, 2017*

What are the chances that it will rain today? Which team has odds in favor in today’s cricket match? What is the probability that Sensex will close above 30K today? All these questions pointed towards the chance, the likelihood of happening the uncertain event. And to determine this chance there’s a technique, a special mathematical and statistical subject entirely related to this known as **Probability**. Probability is nothing else but a chance that some event might occur. More formally, it calculates a numerical value between 0 and 1 that represents the likelihood that an event might occur. It is denoted by **P(E)** where **P** is the probability of event **E **happening. It’s undeniable that probability is a very useful and significant mathematical and statistical tool. The use of this technique is omnipresent. It is required in every field whether science, economics, finance etc. Since, this concept is so essential and significant and as future business managers, we need to understand it and learn it’s application. For the same reason, it has been included in Quantitative Aptitude section of CAT paper. It is an interesting and very simple topic though questions on this topic can at times might be mindboggling and puzzling. But, a thorough understanding can turn the tables and make such questions a child’s play. Today, in this blog, let’s learn some of the useful techniques and methods of probability that will assist you in solving questions during the exam.

To understand probability in its basic form son that solid foundation can be built, in this blog I will explain you using experiments in the simplified form. Before directly moving on to explaining the concept of probability and its varied forms let’s first understand some basic terminology frequently used in its concepts.

**Experiment:** An operation which can produce some well-defined outcomes, is an experiment.

**Random Experiment: **If each trial of an experiment is conducted under identical conditions, the outcome is not unique, but may be of any possible outcome then such experiment is known as random experiment.

**Sample Space: **The set of all possible outcomes in 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, the both **S **and null set φ both are events. **S **is called 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 **forms exhaustive set of events.

**Mutually Exclusive events: **A set of events is called mutually exclusive events if occurrence of one of the them precludes the occurrence of any of the remaining events. In other words, if there are two events A and B then they are called mutually exclusive if AꓵB = φ.

**Independent events: **

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

- The event ‘either A or B’ is said to occur if at least one of A and B occurs. It is usually denoted as AꓴB (A or B).
- The event ‘both A and B’ is said to occur if both A as well B occur simultaneously. It is represented as AꓵB (A and B).
- The event ‘A not’ occurs if A does not occur. It is written as Aᶜ.
- The event ‘A but not B’ is said to occur if A occurs but B does not occur. It is denoted by A-B.

Now as we have understood these some basic terms of probability we can smoothly move forward our area of interest, Probability.

**Probability of an event: **It basically means in 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 probability of the event.

- Probability of the null event is 0.
- 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 A be an event. Then the probability of A is defined as

__No. of favorable outcomes for event A __

**Total no. of outcomes.**

**=n(A)/ n(S).**

For ex**. If an unbiased die is rolled once then what is the probability of getting 3?**

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

Therefore, the chance of getting 3 when die is rolled is = 1 / 6.

We can now head forward to acquaint some different forms and special types of probability. Given below are some important theorems that might help us in formulating solutions to the problems.

- 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 the sample space. Let A and B be any two events. A ≠ φ. Then, probability of the event B, if A have already been occurred, is called conditional probability of B restricted to the occurrence of A. It is represented 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(A****ꓵB)/ n(A) = P(A****ꓵB)/P(A)**

- P(AꓵB) = P(A). P(B/A)

If A ≠ φ & B ≠ φ then,

- P(AꓵB) = P(A). P(B/A) = P(A). P(A/B).

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.

**A card is drawn from the well-shuffled pack of card. What is the probability that the card drawn is of heart given that it is a king?**

Let A = card is of heart and B = card is king

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

Now P(AꓵB) = 1/52

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

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

= ¼

The above example was a very basic one. Questions in exam will be more difficult than this. There could be multiple concepts in one question and conditional probability can be one of them. This concept will also be required in Bayes’ theorem. Thus, it should be understood properly.

**Independent Events: **Two events are said to be independent if the probable occurrence or non-occurrence of any one is not affected by occurrence or non-occurrence of the other i.e. two events A and B are independent if

**P(A/B) = P(A/B****ᶜ) = P(A) **

Or, **P(B/A) = P(B/A****ᶜ) = P(B)**

Or, **P(A****ꓵB) = P(A). P(B)**

Consider this example, **A problem is given to three students whose chances of solving it are ****½, ****⅓, and ****¼ respectively. What is the probability that the problem will be solved?**

To solve this question first,

Let there be three events A, B, and C where

A is the event when student with probability ½ solves the question,

where B is the event when student with probability ⅓ solves the question,

where C is the event when student with probability ¼ solves the question,

Therefore, P(A) = ½, P(B) = ⅓, and P(C) = ¼

Now let Aᶜ, Bᶜ and Cᶜ is the event of not solving question by the students respectively.

- P(Aᶜ) = 1-½ = ½, P(Bᶜ) = 1-⅓ = ⅔, P(Cᶜ) = 1-¼ = ¾

Since we need to find the probability of solving question and determining the probability of not solving question is simpler than that and later we can subtract it from 1 to get the probability of our interest.

Also, clearly in this question probabilities are independent of each other as probability of solving by student 1 does not affect the probability of student 2 or 3 directly or indirectly.

Therefore, P(AᶜꓵBᶜꓵCᶜ) = P(Aᶜ). P(Bᶜ). P(Cᶜ) = ½.⅔.¾ = ¼

Hence, the P [problem will be solved] = 1 – P [none solves the problem]

= 1 – ¼ = ¾

**Note: **Three events A, B, and C are independent if P(AꓵBꓵC) = P(A). P(B). P(C)

** **Three events A, B, and C are pairwise independent if P(AꓵB) = P(A). P(B), P(BꓵC) = P(B). P(C) & P(AꓵC) = P(A). P(C)

**Relation between Independent and Mutually Exclusiveness of 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 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. Clearly, former talks about events in an experiment and latter explain independency among multiple experiments.

i.e. (i) consider the tossing of a coin twice, clearly the sample the sample space for the second is not affected by the results of the first toss. So, two tosses are random experiments.

**Bayes’ Theorem: **Suppose A₁, A₂, … A_{n}, are mutually exclusive and exhaustive set of events. Thus, they divide the sample into n parts and the event B occurs. Then the conditional probability that A_{i }happen given that B has happened is given by

P(A/B) **= P(A_{i}). P(B/A_{i})**

__∑ ^{n}_{i=1 }__P(A

Again, to understand the application of above theorem more precisely, consider this example,

**In certain day care class, 30% of children have grey color eyes, 50% have blue color eyes and 20% have other color eyes. One day they play a game together. In the first run 65% of grey color ones, 82% of blue color ones and 50% of children of other eye color was selected. Now if a child is selected is randomly from the class and we know that he/she was not in the first game, what is the probability that the child has blue color eyes.**

As given in the theory, Let A_{1, }A_{2} and A_{3} be the events that the student has eyes color grey, blue, and other respectively.

Therefore, P(A_{1}) = 0.3, P(A_{2}) = 0.5, P(A_{3}) = 0.2

Now it’s been in question that students are randomly selected to play the game. Let E be the event of selecting a student for game 1.

Therefore. E’ be the event of selecting student in game 2.

Thus, P(^{E’}/_{A1}) = 1 – 0.65 = 0.35; P(^{E’}/_{A2}) = 1 – 0.82 = 0.18; P(^{E’}/_{A3}) = 1 – 0.5 = 0.5

We have to find the probability that the selected child in game two has blue eyes i.e. P(^{A2}/_{E’})

Using Bayes’, we get,

P(^{A2}/_{E’}) = (0.5*0.18) / 0.3*0.35 + 0.5*0.18 + 0.2*0.5 = 0.305

There’s one more important concept of statistics which is frequently used to simplify the calculation and it gives a straightforward solution. It is a probability distribution known as Binomial Distribution.

**Binomial Distribution for successive Events: **Suppose p and q are the respective chances of the happening and failing of an event at a single trial. Then the chance of its happening r times in trials is **^{n}C_{r}p^{r}q^{n – r} **because the chance of its happening r times and falling n – r times in given order is

**A fair dice is tossed repeatedly until six shoes up three times. The probability that exactly 5 tosses are needed will be?**

Now the probability of getting ‘six’ on the dice in one trial is ⅟_{6. }Therefore ‘p’ here is ⅟_{6 }and thus, q will be ^{5}/_{6. }We need six three times in exactly 5 trials. So, let’s fix that in 5^{th }6 shows up and its probability will be ⅟_{6}. Hence the probability of getting six two times out of the 4 trials using the above distribution will be = ^{4}C_{2} *(^{5}/_{6})^{2} * (⅟_{6})^{2}.

Therefore, the probability of getting three ‘six’ in exactly 5 tosses is [ ^{4}C_{2} *(^{5}/_{6})^{2} * (⅟_{6})^{2}] * ⅟_{6} = 25/1296.

The above problem was just one example of the use of Binomial Distribution. Its application is wide and generally used in many types of problems. There are many other distributions too but those are not relevant in the purpose of CAT. But, in exam questions based on conditional probability, Bayes’ or amalgamation of all the above. Dice and coin toss problems with complications can be asked in the exam so you should have the thorough practice of these.

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All questions from CAT 2017 Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is

Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?

Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Quantitative Aptitude – Modern Maths – P&C – Q5

You can also see **Cube Based Puzzles in Logical Reasoning for CAT Exam Preparation**

*This article was contributed by Sejal Khurana. If you want to write for us, please email us on [email protected]*

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