Probability

Probability is a very popular topic of the Mathematics syllabus of the IIT JEE. It is not considered to be a very simple topic, but it is important to master the topic in order to remain competitive in IIT JEE. Majority of the stduents find it difficult to master this topic because their basics are not clear. In this chapter, we shall study the methodologies of computing the probability of an event, be it complex or simple. We shall try to comprehend the tools used for this purpose. In an attempt to make the subject easily comprehensible and clearly understandable a large number of illustrations and problems are selected and solved. Every aspect is fully supported by sufficient number of relevant examples. The students will note that for the understanding of the subject matter of this chapter, the knowledge of permutation and combination is essential. 

We shall first throw some light on the meaning of probability and then proceed toward the various terms related to it.

What do you mean by the term probability?

In simplest possible words, probability is the degree of likeliness of the happening of an event. It is not always possible to predict an event with total certainity. In such circumstances, the concept of probability is used to predict how likely are they to happen or not. It may also be defined as a measure or estimation of likelihood of occurrence of an event. Probabilities are given a value between 0 (0% chance or will not happen) and 1 (100% chance or will happen). The higher the degree of probability, the more likely the event is to happen, or, in a longer series of samples, the greater the number of times such event is expected to happen.We disucss this in detail with the help of following examples:

1. Throwing a dice: When we throw a dice, there can be six possible outcomes. We can get any number among 1 to 6 and hence the total number of possible outcomes is 6. The probability of getting any number from 1 to 6 is therefore 1/6.  But what is the probability of getting a 7 on the throw of a dice?  Since it is not possible to get a 7 on a dice hence, its probability is zero.

 

2. Tossing a coin: When a coin is tossed, we can either get a head or tail. So, there can be two possible outcomes and the rpobability of each is ½. 

Hence, P(H) = P(T) = ½.

 

 

It is clear form the above examples that probability of an event just denotes the possibility of its occurence or non-occurence. Hence, we get the following formula:

Probability of occurence of event A = P(A) = Number of favourable outcomes/ Total number of outcomes   

 

Remark: 

1. The number of outcomes favourable to A are denoted by n(A) and total number of outcomes in sample space are denoted by n(S). Hence, the above formula becomes P(A) = n(A)/n(S).

2. The probability of an event can vary between 0 to 1, i.e. 0 ≤ p ≤ 1.

3. Probability can never be negative.

4. Probability of occurence of an event = 1 – (Probability that it doesn’t occur).

 

Illustration 1:

A glass jar contains 5 red, 6 green, 3 blue and 8 yellow coins. If a single coin is chosen at random from the jar, what is the probability of choosing a red coin, a green coin, a blue coin or a yellow one?

Solution:

The possible outcomes of the experiment are red, green, blue and pink. Using the above mentioned formula, we have

P(red) = Number of ways to chose red/ total number of coins 

          = 5/22

P(green) = Number of ways to chose red/ total number of coins 

          = 6/22

P(blue) = Number of ways to chose red/ total number of coins 

          = 3/22

P(yellow) = Number of ways to chose red/ total number of coins 

          = 8/22

This shows that the outcomes are not equally likely.

Refer the following video for more information on Probability.

 

The various topics covered in this chapter are listed below:

  • Experiment,Sample Space and Event

  • Introduction to Probability

  • Conditional Probability

  • Independent events

  • Bayes’ Theorem

  • Binomial Distribution for successive events

  • Solved Examples

All the above topics have been discussed in detail in the coming sections. Here, we shall just give a brief outline of each of them:

Experiment or Trial:

An experiment or a trial refers to an action whose result is not certain. Some examples of trials include tossing a coin, throwing dice, drawing a card from a deck of cards.

Sample Space:

All the possible outcomes of an experiment are known as the sample space. 

Eg: There are 52 cards in a deck. Hence, the sample space is all 52 cards. 

Sample Point:

A sample space is amde up of sample points whihc refers to just one of the possible outcome. 

Eg: In a deck of cards, getting the King of Hearts is a sample point. Please note that "King" is not a sample point as there are 4 Kings which constitute 4 different sample points.

Obtaining a ‘2’ when a dice is thrown is a sample point.

Event:

A single result of an experiment is called an event. Some examples of events include getting a tail on the toss of a coin is an event.

We clarify the difference between these terms with the help of the following figure

Independent events:

Two events are said to be independent of each other if the probability of the occurrence of one event does not affect the probability of the occurrence of the other event.

Example: Suppose you rolled a dice and tossed a coin. Now the probability of getting a particular number on the dice in no way influences the probability of getting a head or a tail on the coin. 

Remark:

Three events A, B and C are independent iff all the given conditions hold good:

P(A ∩ B) = P(A) . P(B), P(B ∩ C) = P(B) . P(C), P(C ∩ A) = P(C) . P(A)

P(A ∩ B ∩ C) = P(A).P(B).P(C) 

Hence, three events are said to be independent if they are independent in pairs as well as mutually independent.

Similarly, if we have n events A1, A2, …. An, then they are said to be independent if they satisfy nC2 + nC3 + …. + nCn = 2n – n – 1 conditions. 

Conditional Probability: 

The conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written P (B|A), read as the probability of B given A. In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is P(B).

If events A and B are not independent, then the probability of the intersection of A and B (the probability that both events occur) is defined by P (A and B) = P (A) P (B|A).

From this result, we also derive the following result

P (B/ A) = P (A and B) / P (A)

Note: For mutually exclusive events, P(A/B) = 0

Bayes Theorem:

If an event A can occur with one of the n mutually exclusive and exhaustive events B1, B2, … Bn, and the probabilities P(A/B1), P(A/B2), … P(A/Bn) are known, then   

Note:

P(A ∩ Bi) = P(A). P(Bi/A) = P(Bi).P(A/Bi)

So, P(Bi/A) = P(Bi).P(A/Bi)/ P(A) = P(Bi).P(A/Bi)/ ΣP(A∩Bi)

Binomial Distribution for Successive events:

Binomial distribution is one of the most important distributions in the probability theory. The probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions is called the binomial distribution. 

Let probability of success in any trial be p and that of failure be q, then 
p + q = 1

Then (p + q)n = C0 Pn + C1Pn-1q +...... Crpn-rqr +...+ Cnqn

Then the probability of exactly k successes in n trials is given by

Pk = nCkqn-kpk

Assumptions of Binomial Distribution:

1. The number of observations n is fixed.

2. It gives one of the two possible outcomes, say 'success' (when required event take place) and failure (when required event does not take place)

3. Each trial has the same probability of success.

4. Each trial is mutually exclusive.

5. The probability of getting at least k successes is P(x > k) = Σnx=k  nCx px qn-x

6. Σnx=k  nCx qn-x px = (q + p)n = 1.

7. Mean of binomial distribution is np.

8. Variance is npq,

9. Standard deviation is given by (npq)1/2, where n,p and q have their meanings as described above.

Important assumptions have been depicted below in the form of chart:

 

Illustration 2:

A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is almost 0.3. Is it possible that the probability of B getting selected is 0.9?(IIT JEE 1982)

Solution:

Let P (A) and P (B) be the probabilities of selection of A and B respectively.

P (A) = 0.5, P (A∩B) ≤ 0.3

P(A∪B) = P(A) + P(B) – P(A∩B) ≤ 1

P(B) ≤ 1 + P(A∩B) – P(A)

          ≤ 1 + 0.3 – 0.5 ≤ 0.8

Hence, the probability of selection of B cannot be 0.9.

 

Illustration 3:

A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. Find the probability of occurrence of A.

Solution:

It is given that P (A).P (B) = 1/6

And also P (Ac) P (Bc) = 1/3

This can also be written as [1-P (A)]. [1-P (B)] = 1/3

Let P(A) = x and P(B) = y

Then (1-x)(1-y) = 1/3 and xy = 1/6

Hence, 1-x-y+xy = 1/3 and xy = 1/6

x+y = 5/6 and xy = 1/6

x(5/6 – x) = 1/6

6x2 – 5x + 1 = 0.

(3x -1) (2x-1) = 0

So x = 1/3 and 1/2

Hence, the probability of A i.e. P(A) = 1/3 or 1/2.

 

Illustration 4:

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied it is 1/8. Find the probability that he knew the answer to the questions given that he correctly answered it.

Solution:

let us use the following symbols for denoting the various options

G for guesses

C for copies

K denotes the possibility that the examinee knows

R if the answer is right

So, P(G) = 1/3

P(C) = 1/6

P(K) = 1- (1/3 + 1/6) =1/2

Now R = (R∩G) ∪ (R∩C) ∪ (R∩K)

P(R) = P(G) P(R/G) + P(C) P(R/C) + P(K) P(R/K) … (1)

Now, P(R/G) = 1/4

P(R/C) =1/8

P(R/K) =1

Putting this in equation (1), we obtain

P(R) = 1/3. 1/4 + 1/6.1/8 + 3/6. 1

         = 1/12 +1/48 + 3/6

         = 29/48

Hence, the required probability = P(K∩R)/ P(R)

                                             = 24/29

 

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