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

Conditions for the application of Baye's formula is that priori events i.e. A₁, A₂, ......., A_n of the sample space are exhaustive and mutually exclusive i.e.

         A₁ U A₂ U ........... U A_n = S

        and   A_i ∩ A_j = Φ j, i = 1, 2, ........ n and i ≠ j

        It's derivation is as follows

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

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

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

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

        Hence, P(B_j/A) = = (P(B_j ).P(A/B_j) )/(∑ⁿ_(i=1)  P(B_i ).P(A/B_i))

        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 Baye's Theorem

Note:  To recognize the question in which Baye's theorem is to be used, the key word is " is found to be".

 Illustration:

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

Solution:

        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 is turned up this is an information to us. You should revise the probability of priori events A_i 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 Baye's formula

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

        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

Note:This problem is very simple and illustrated only to make the application of Baye's formula clear.

Illustration:

In a factor, 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?

Solution:

        In the formula P(A_k/A) = (P(A_k )P(A/A_k))/(∑ⁿ_(i=1) P(A_i ).P(A/A_i))

A_i means all the possibilities, which can happen w.r.t. to the given event A, while A_k means the particular event whose probability w.r.t. 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(A_i ).P(A/A_i))

       =>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.

Illustration:

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

Solution:

First of all let's try to find out all the possibilities, which can happen 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 ( A )

Proof: This is so because

P(A_i/B)= P(A_i ∩ B)/P(B) = P(A_i ∩ B)/P((B ∩ A_i) U P(B ∩ A₂) U...(B ∩ A_n)) = P(A_i ∩ B)/P((B ∩ A₁) U P(B ∩ A₂) U...(B ∩ A_n))

= P(Ai).P(B/Ai)/∑ⁿ_i=1  P(A_i)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.

Illustration:

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.

Solution:

Here A₁, A₂, A₃ are the events that the bag 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), 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 .

Illustration:


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?


Solution:

Here, the 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/1_w) -> Probability that B occurs when exactly 1 w ball is there = ¹C₁/⁵C₁ and so on and 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 )] )


= 1/(0+1+2+...+5)=1/15.