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Bayes’ Theorem

Bayes theorem is an important topic of probability and it fetches many questions in IIT JEE. This topic is very interesting and can be easily understood with the help of real life examples.

Most of the things happening in day to day life are conditional. If we consider the question, “What is the life expectancy of a male?” The probable answer could be somewhere around 72 years. But suppose the person asking this question is 55 years old then a better question could be, “What is the life expectancy of males who have lived 54 years?” In this case, the answer would obviously be more than 72. Thus, in this case we wish to rule out an estimate “conditioned on” the fact that a person has already lived 55 years. Life is not constant and is constantly changing. Majority of the decisions and results are based on invariable information. This is where the Bayes theorem has an edge over others. This theorem utilizes dynamic information to give a more accurate answer.

Let us consider another example. A headline once said that married men live longer than single males. Assume that the article says that a married male’s expected life is around 72 years while for a single male, its 66 years. Various reasons considered were, a better diet offered by wife, no to smoking and drinking etc. Bayes theorem offers a solution to this- how many married males died at the age of two? Instead, if the article said that the life expectancy of married as 55 and of single as 55 years, then the numbers would almost be the same.

Hence, the Bayes theorem gives an accurate mathematical representation which helps in estimating Prob (Condition A | Condition B), which is read as “probability of condition A given that condition B already exists or has occurred.”

Bayes Theorem for IIT

Bayes’ theorem revises (reassigns) the probabilities of the events A1, A2, .........., An, related to a sample space, when there is an information about the outcome beforehand. The earlier probabilities of the events P (Ai), i = 1, 2, ............, n are called a priori probabilities and the probabilities of events calculated after the information A is received i.e. P (Ai /A) is called posterior probabilities.

Conditions for the application of Bayes’ formula are:

  • Priori events i.e. A1, A2, ......., An of the sample space are exhaustive and mutually exclusive i.e.

A1 U A2 U ........... U An = S

and   Ai ∩ Aj = Φ j, i = 1, 2, ........ n and i ≠ j

  • Within the sample space there would exist an event B such that P (B)>0.

  • The main aim is to compute a conditional probability of the form P (Ai /B).

  • We know that at least one of the two sets of the two probabilities are given below:

(i)                P(Ai ∩B) for each Ai

(ii)             P(Ai) and P(B/Ai) for each Ai

 Its derivation is as follows

Let B1, B2, ............, Bn be n mutually exclusive and exhaustive events and A be any event in sample space, then

A = (A ∩ B1) U (A ∩ B2) U ......... (A ∩ Bn)

=> P(A) = P(A ∩ B1) + P(A ∩ B2) + P(A ∩ B3) ........+ P(A ∩ Bn)

= P(B1).P(A/B1) + P(B2).P(A/B2) +.........+ P(Bn).P(A/Bn)

Hence, P(Bj /A) = = (P(Bj ).P(A/Bj) )/(∑n(i=1)  P(Bi ).P(A/Bi))

Note:  If in a problem some event has already happened and then the probability of another event is to be found, it is an application of Bayes Theorem. To recognize the question in which Bayes’ theorem is to be used, the key word is “is found to be".


A die is rolled and it is found that number turned up is an even number. Find the probability that it is 2.


        All possible events when we roll a die are

                A1 : 1 appears P(A1) = 1/6

                A2 : 2 appears P(A2) = 1/6

                A3 : 3 appears P(A3) = 1/6

                A4 : 4 appears P(A4) = 1/6

                A5 : 5 appears P(A5) = 1/6

                A6 : 6 appears P(A6) = 1/6

An even number has turned up this is an information to us. You should revise the probability of priori events Ai in the light of information received. (It will be totally foolish if we don't revise the probabilities. Since the information of even numbers is available, it makes the probabilities of 1, 3, and 5 equal to zero).

 Let A: Even number has turned up.

We have to calculate the probability of A2 when A is given. Since A1, A2, .........., A6 are exhaustive and mutually exclusive we can apply Bayes’ formula

P(A2/A) == (P(A)P(A/A2))/(P(A1 )P(A/A1)+P(A2 )P(A/A2)+.……+P(A6 )P(A/A6))

where P(A/A1) means probability of A when A1 is given i.e. probability of coming of an event number when 1 has appeared. Obviously it is zero.

Hence P(A/A1) = 0

Similarly, P(A/A2) = 1, P(A/A3) = 0, P(A/A4) = 1, P(A/A5) = 0, P(A/A6) = 1

Therefore, P(A2/A) = (1/6×1)/(1/6×0+1/6 ×1+1/6×0+1/6×1+1/6×0+1/6×1)=1/3

Hence this is the required probability.

Note: This problem is very simple and illustrated only to make the application of Bayes’ formula clear.


In a factory, machines A, B and C manufacture 15%, 25% and 60% of the total production of bolts respectively. Of the bolts manufactured by the machine A, B and C 4%, 2% and 3% are defective A bolt is drawn at random and is found to be defective. What is the probability that it was produced by B?


In the formula P(Ak/A) = (P(Ak )P(A/Ak))/(∑n(i=1) P(Ai ).P(A/Ai))

Ai means all the possibilities, which can happen w.r.t. to the given event A, while Ak means the particular event whose probability w.r.t. the event B we are required to find

Let us take A = the event of bolt being defective.

A1 = the bolt is produced by B.

A2 = the bolt is produced by A.

A3 = the bolt is produced C.

Required probability = P(A1/A)=(P(A1 )P(A/A1))/(∑3(i=1) P(Ai ).P(A/Ai))

=>P(A1/A) =(25/100×2/100)/((15/100×4/100)+(25/100×2/100)+(60/100×3/100))

= 50/(60+50+180)=5/29.

You may also refer the video for more examples on Bayes’ Theorem 


A bag contains 5 balls of unknown colors two balls are drawn at random and are found to be red. Find the probability that the bag contains exactly 4 red balls.


First of all let's try to find out all the possibilities, which can occur w.r.t. to given events. Since two balls drawn are found to be red, so there are four possibilities. The bag contains only two red balls (say event A2), bag contains 3 red balls (A3), bag contains 4 red balls (A4), or that all the balls are red balls (A5).

Let B: two balls are drawn and found to be red. So, it is obvious that we have to find the probability of A4 given B i.e. required probability = P(A4/B)


Now as all these four possible cases are equally likely i.e. we cannot say that bag is more likely to contains 5 red balls, than 3 red balls, so priori probabilities of all these events will be equal i.e.

P(A2) = P(A3) = P(A4) = P(A5)

Since A2, A3, A4 and A5 are exhaustive

.·.      P (A2) + P (A3) + P (A4) + P (A5) = 1

.·.     P (A4) = ¼ (since A2, A3, A4 and A5 are equally likely also)

Now P (B/A4) means that bag contains 4 red balls and we have to find the probability, that two balls drawn at random are red which is obviously 4C2/5C2 .

So P(A4/B)=(1/4×(4C2)/(5C2))/(1/4×(2C2)/(5C2)+1/4×(3C2)/(5C2)+1/4×(4C2)/(5C2)+1/4×(5C2)/(5C2))

 = 12/39

Proof: This is so because

P(Ai/B)= P(Ai ∩ B)/P(B) = P(Ai ∩ B)/P((B ∩ Ai) U P(B ∩ A2) U...(B ∩ An)) = P(Ai ∩ B)/P((B ∩ A1) U P(B ∩ A2) U...(B ∩ An))

= P (Ai).P (B/Ai)/∑ni=1 P (Ai) P (B/A1)

Note: In conditional probability the sample space is reduced to the set of samples/outcomes in the event which is given to have happened.


A rare genetic disease is discovered. Although only one in a million people carry it, you consider getting screened. You are told that the genetic test is extremely good; it is 100% sensitive (it is always correct if you have the disease) and 99.99% specific (it gives a false positive result only 0.01% of the time). Having recently learned Bayes' theorem, you decide not to take the test. Why?


Bayes' Theorem states that for events X and Y:

P (X|Y) =P (Y|X)*P(X)/P(Y).

We want to know the probability of being healthy(X) given the positive test (PT) results(Y).

According to the Bayes' Theorem,

P (healthy|PT) =P (PT|healthy)*P (healthy)/P (PT).

From the problem we know that

P (healthy) =1-0.000001=0.999999

And getting a false positive

P (PT|healthy) = 0.0001.

The only unknown in the formula above is the probability of having a positive test P (PT). It can be calculated using the definition of marginal probability


where Zi, i=1...n are all possible events. In our case there are only two possible events: "being healthy" and "being sick". Therefore

P (PT) =P (PT|healthy)*P (healthy) +P (PT|sick)*P (sick).

From the problem we know that

P (PT|sick) =1.0
(Test is always correct in presence of the disease) and

P (sick) =0.000001.

Substituting the numbers into the formula we get



P (healthy|PT)=0.0001*0.999999/0.000101=0.990098,
that is very close to 1.
So, the probability of still being healthy given that the results of the test turned positive is above 99%. That is a good reason for not taking the test.


An urn B1 contains 2 white and 3 black balls and another urn B2 contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was B1.


Let E1, E2 denote the events of selecting urns B1 and B2 respectively.

Then P (E1) = P (E2) = 1/2

Let B denote the event that the ball chosen from the selected urn is black.

Then we have to find P (E1/B).

Also, probability of selecting a black ball from the urn 1 i.e. P (B/E1) is 3/5.

Probability of selecting a black ball from the urn 2 i.e. P (B/E2) is 4/7.

Hence the required probability i.e. P(E1/B) = P(E1) P(B/E1) / [P(E1) P(B/E1)+ P(E2) P(B/E2)

= ( ½. 3/5 )/ ½.3/5 + ½. 4/7

= 21/41. 


Each of three bags A, B, C contains white balls and black balls. A has a1 white & b1 black, B has a2 white & b2 black and C has a3 white & b3 black. A ball is drawn at random and is found to be white. Find the respective probability that it is from A, B & C.


Here A1, A2, A3 are the events that the bags picked are A, B, C respectively

E is the event that a white ball is drawn.

We are supposed to find P (A1/E), P (A2/E), and P (A3/E).

P(A1/E) = P(A1UB)/P(E) = ((Prob. that bag A is chosen and white is drawn)/(Prob. that a bag is chosen at random and white is drawn))

= (P(A1 ).P(E/A1 ))/(P(A1 ).P(E/A1 )+P(A2 ).P(E/A2 )+P(A3 ).P(E/A3 ) )

= (1/3.a1/(a1+b1 ))/(1/3.[a2/(a2+b2 )+a2/(a2+b2 )+a3/(a3+b3 )] )=p1/(p1+p2+p3 )  

Similarly, P (A2/E) = P2/P1+P2+P3 , P(A3/E) = P3/P1+P2+P3

where p1 = a1/a1+b1 ,          p2 = a2/a2+b3 ,           p3 = a3/a3+b3 .


A bag contains 5 balls and of these it is equally likely that 0, 1, 2, 3, 4, 5 are white. A ball is drawn and is found to be white. What is the chance that is the only white ball?


Here again, is a problem of conditional probability.

The condition B that is given is that one ball is drawn and is white.

Hence P(1w/B)=(P(1w/B).P(1w))/(P(0w)P(B/0w)+P(1w),p(B/1w)+..+P(5w).P(B/5w))

Where P(B/1w) - Probability that B occurs when exactly 1 w ball is there = 1C1/5C1 and so on

P(1w) = P(2w) =………= P(5w) = P(0w) = 1/6 

=> the required probability = (1/6 (5 C1)/(5 C1))/(1/6 [0+(1 C1)/(5 C1)+(2 C1)/( 5C1 )+...( 5C1)/( 5C1 )] )=


You may also consult the Previous Papers to get an idea about the types of questions asked.


Find P (M/N), given P (N/M') = 0.7, P (M) = 0.2, and P (N/M) =0 .8.

Solution: P (M') = 1 - P (M) = 0.8.
By Bayes' Theorem,

P(M'/N)/P(M/N) = (P(N/M')P(M')/P(N))/(P(N/M)P(M)/P(N)) = 0.7*0.8/(0.8*0.2) = 7/2.
Hence 1 = P(M/N) + P(M'/N) = P(M/N)(1 + P(M'/N)/P(M/N)) = P(M/N)(1 + 7/2),
and so P(M/N) = 1/(1 + 7/2) = 2/9.


Find P (M'/N), given P (N/M') = 0.4, P (M) = 0.6, and P (N/M) =0.2.

Solution: P (M') = 1 - P (M) =0.4.
By Bayes' Theorem, P(M'/N)/P(M/N) = (P(N/M')P(M')/P(N))/(P(N/M)P(M)/P(N)) = .4*.4/(.2*.6) = 4/3.

Hence 1 = P(M/N) + P(M'/N) = P(M'/N)(P(M/N)/P(M'/N) + 1) = P(M'/N)(3/4 + 1),
and so P(M'/N) = 1/(3/4 + 1) = 4/7.

Bayes’ Theorem is an important section of Probability. It is a simple and an interesting topic and fetches 2-3 questions in the exam. With a bit of hard work the topic can be mastered easily and it is important to excel in this topic to remain competitive in the IIT JEE.

To read more, Buy study materials of Probability comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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