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**In this Module of Probability you will be covering following Objectives**

- Probability
- Meaning & definitions of Probability
- Theorems
- Methods & shortcuts to solve the problems

**Text - Comments**

Probability is one of the most important mathematical concepts that has a real presence & impact in our daily lives. There is a consistent presence of probability questions in CAT & other B school entrances at a min of 2 questions every year. This makes this chapter extremely important for an aspirant. The students who didn't have Mathematics as their core subject after Class X/XII th classes have been seeing CAT exam as a test of Class XII standard. This leads to students taking a negative approach while tackling/preparing for the Quantitative aptitude section. EduRev suggests Students that the Math asked in MBA entrance is mainly logical while studying the chapter.

In this chapter you intend to improve your concepts of probability to such an extent that you feel in total control of this topic through easy methods of explanations & practice examples. For those who are experienced in the topic, you advise you to use this chapter both for revising the basic concepts and practice the examples.

**What is Probability?**

Probability is simply how likely something is to happen. Whenever you’re not certain about the outcome of an event, you can talk about the probabilities of certain outcomes—how likely they are.**Ex:**

- The best example for understanding probability is flipping a coin
- Probability (Chance) of rain on a given day.
- Dice outcomes

**Application of Probability**

While you move into the application & problems in probability, one question that strikes in every question of probability is that whenever one is asked the question, what is the probability? the immediate question that arises/should arise in one’s mind is the probability of what?

The answer to this question is the probability of an EVENT. The EVENT is the cornerstone or the bottomline of probability. Hence, the first aspect of solving a question on probability is to define the event.

Lets see an example,**Example ****Ques. Imagine an instance where you are throwing a dice, what is the probability of getting a number greater than 3, in a throw of a normal unbiased dice having 6 faces?Sol.**

The

Hence, the required probability is 1/6 + 1/6 + 1/6 = 3/6 = 1/2.

Can you try this question?

**⅓****⅔****⅚****⅖**

**Sol.**** **Here the answer is ⅚ ,

We have the event of getting any number greater than 1, you have following possibilities for such an event - 2. 3. 4. 5 & 6.

The individual probabilities of each of these are ⅙,⅙, ⅙ ,⅙ and ⅙ respectively.

Hence, the required probability is ⅙ + ⅙ + ⅙ +⅙ + ⅙ = ⅚.**Usually you would encounter two major types of problems in Probability with the the use of conjunctions AND and OR:**

In general, event definition means breaking up the event to the most basic building blocks, which are commonly through the two English conjunctions— AND and OR.**‘AND’ as a binding conjunctive.**

Whenever you see AND as the natural conjunction joining two separate parts of the event definition, you can replace the AND by the multiplication (X) sign.

Thus, if E AND R have to occur, and if the probability of their occurrence are P(E) and P(R) respectively, then the probability that E AND R occur is obtained by connecting P(E) AND P(R). Replacing the AND by multiplication sign you get the required probability as:

Req Probability = P(E) X P(R)**Example. Let’s try solving an example, If you have the probability of E hitting a target as 1/2 and that of R hitting the target as 1/5, then the probability that both hit the target if one shot is taken by both of them is?Sol.** Event here is that E hits the target AND R hits the target, both hitting the target. Hence the resulting probability is going to be

Probability = P(E) AND P(R) = P(E) X P(R)

Using the values given of P(E) & P(R),

P(E) X P(R) = ½ X ⅕ = 1/10

**EduRev Tip:** When both events are connected by ‘AND’, please remember to use multiplication of events. (Note: Remember the above explanation, where you used Multiplication when both events are connected by conjunctive AND. That is the reason why you multiplied them.)**Practice Question**

Try yourself:**Ques. In a throw of two dice, find the probability of getting one prime and one composite number.**

View Solution

Sol.

Probability = P(E) OR P(R) = P(E) + P(R)

Using the values given of P(E) & P(R),

P(E) + P(R) = ½ + ⅕ = 7/10

Unlike above example where the event will happen only with both E & R events occurring , here if any one of E, R events can occur.

**EduRev Tip:** Look for the language of question, conjunction ‘OR’ is used to connect the events, which means that you need to add them up.**Combination of ‘AND’ and ‘OR’**

You would encounter where in once the event is defined, the probability of each sub event within the broad event is calculated and all the sub events are connected by Multiplication (for AND) or by Addition (for OR) to get the final solution.**Example. If two dice are thrown, what is the chance that the sum of the numbers is not less than 10.Sol. **

Which can be done by

P(E) = (6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR (5 AND 6) OR (6 AND 6) that is, ⅙ X ⅙ + ⅙ X ⅙ + ⅙ X ⅙ + ⅙ X ⅙ + ⅙ X ⅙ + ⅙ X ⅙ = 6/36 = ⅙

Hence for the event of sum less than 10, P(E’) = 1 - P(E) = 1 - ⅙ = ⅚ .

can see that no matter how many broad & sub events the problem of probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.

Try yourself:**There are two bags containing white and black balls. In the first bag, there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black.**

View Solution

Try yourself:**In a four game match between Monica and Chandler, the probability that Chandler wins a particular game is 2/5 and that of Monica winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the match is drawn (Score is 2–2).**

View Solution

**EduRev Tip: ** In the above question you tried to identify the sub-events & try to add them based on the event definition.**Important Consideration during defining Events****:**

Brief Intro,

While attempting the questions, you need to take into considerations certain activities/events. Questions in CAT exams usually have direct or indirect mention of such events.**Impossible event and certain event**

An impossible event has no chance of occurring. If event A is impossible, then P(A) = 0

A certain event is certain to occur. If event A is certain, then It is usually represented by P(A) & the value of P(A) = 1.**Random experiment** An experiment whose outcome has to be among a set of events that are completely known but whose exact outcome is unknown is a random experiment (e.g. Throwing of a dice, tossing of a coin). Most questions on probability in CAT are based on random experiments.**Sample space** This is defined in the context of a random experiment and denotes the set representing all the possible outcomes of the random experiment. [e.g. Sample space when a coin is tossed is (Head, Tail). Sample space when a dice is thrown is (1, 2, 3, 4, 5, 6).]**Mutually Exclusive Events**

Two or more events are said to be mutually exclusive; these events cannot occur simultaneously. Two or more events are said to be compatible if they can occur simultaneously. Two events (A and B) are mutually exclusive if the intersection of two events is null or they have no common element i.e. A ∩ B = φ.

e.g. In drawing a card from a deck of 52 cards:

A: The event that it is red.

B: The event that it is black.

C: The event that it is a king.

In the above case events A and B are mutually exclusive but the events B and C are not mutually exclusive or disjoint since they may have common outcomes.** Equally likely events:** If two events have the same prob- ability or chance of occurrence they are called equally likely events. (In a throw of a dice, the chance of 1 showing on the dice is equal to 2 is equal to 3 is equal to 4 is equal to 5 is equal to 6 appearing on the dice.)

Required numbers are 11, 22, - - - - - 99 ⇒ 9

So probability = 9/90 = 1/10

The set of all possible outcomes of a random experiment is known as the

P(3) = 1/6

P(head and 3) = 1/2.1/6 = 1/12

If there are n-elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening of A is denoted by P (A) and is defined as the ratio m/n.

Probability of an event occurring =

Thus, P (A) = m/n

Clearly, 0 ≤ m ≤ n, therefore 0 ≤ m/n ≤ 1, so that 0 ≤ P (A) ≤ 1

**Sum Rule**

If E and F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by:**P(E or F) or P (E ∪ F) = P(E) + P(F)**

If the event is not mutually exclusive, then**P(E ∪ F) = P(E) + P(F) – P(E and F together)****Note:** P (neither E nor F) = 1 – P(E or F).**Complement of an event**

Since the number of cases in which the event A will not happen is n - m, therefore ifdenotes not happening of A, then the probability P () of not happening of A is given by

P () = (n-m)/n = 1 – m/n = P (A) or P(A) + P () = 1**Example. From a bag containing 8 green and 5 red balls, three are drawn one after the other. Find the probability of all three balls being green if(a) the balls drawn are replaced before the next ball is picked(b) the balls drawn are not replaced.Sol.**

Hence answer to the question above will be:

Required probability = 8/13 X 8/13 X 8/13 = (83/133)

(b)When the balls are not replaced, the probability of drawing any color of ball for every fresh draw changes. Hence, the answer here will be:

Required probability = 8/13 X 7/12 X 6/11

**EduRev Tip:** If n dice are thrown then total possible outcomes = 6n & if n coins are tossed then possible total outcomes = 2n

**Multiplication Rule**

When two events, A and B, are independent, the probability of both occurring is:**P(A and B) = P(A ∩ B) = P(A) × P(B)****Binomial****Example. A dice is tossed 5 times. What is the probability that 5 shows up exactly thrice?****Sol.** Here the 'random experiment' consists of tossing a die 5 times and observing the number '5' as success, then p = Probability of getting '5' with single dice = 1/6 so that q = 1-1/6 = 5/6

Since the value of p is constant for each dice and the trials are independent, using formula for Binomial probability law, the probability of r successes is given by :

Required probability that shows up exactly thrice is given by:**Practice Question**

Try yourself:An unbiased dice is thrown. What is the probability of getting

(i) an even number;

(ii) a multiple of 3;

(iii) an even number or a multiple of 3;

(iv) an even number & a multiple of 3?

Choose the most appropriate option.

(i) an even number;

(ii) a multiple of 3;

(iii) an even number or a multiple of 3;

(iv) an even number & a multiple of 3?

Choose the most appropriate option.

View Solution

Sol.

(i) an ace,

(ii) red,

(iii) either red or king,

(iv) red and a king

Sol.

Therefore exhaustive number of cases = 52C1 = 52.

(i) There are four aces in a pack of 52 cards, out of which one can be drawn in 4C1.

Therefore Favourable number of cases = 4C1 = 4.

So, required probability = 4/52 = 1/13

(ii) There are 26 red cards, out of which one red card can be drawn in 26C1 ways. Therefore,

Favourable number of cases = 26C1 = 26.

So, required probability = 26/52 = 1/2

(iii) There are 26 red cards including 2 red kings and there are 2 more kings. Therefore, there are 28 cards, one can be drawn in 28C1 ways.

Therefore favourable number of cases = 28C1 = 28.So, required probability =

(iv) There are 2 cards, which are red and king. Therefore, favourable number of cases = 2C1 = 2.

So, required probability = 2/52 = 1/26

Amit must throw either 11 or 12.

This can be done in 3 ways namely (6 + 5, 5 + 6, 6 + 6)

Thus Amit’s chance of throwing a higher number is 3/36 or 1/12.

With this you have reached the end of Probability notes of EduRev. Now we suggest you following paths

**V****ideo Link**If there any clarifications you would need regarding concept, please go through the video**Videos links**- https://youtu.be/eUUENwCXJNo
- https://youtu.be/FB6R1qBYHXs
- https://youtu.be/hMVzDx0yQj0?t=86
**Test**Analyse your conceptual understanding by attempting tests**Examples & Tips document**You can revise the whole concept & practice few more examples in these documents.

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