Total Probability Theorem And Bayes' Theorem

TOTAL PROBABILITY THEOREM AND BAYES' THEOREM

Partition of Sample-Space:

Consider a sample space 'S'. Let A₁, A₂, L A_n be the

set of mutually exclusive and exhaustive set of

sample-space S.

These A₁, A₂, L, A_n events are said to partition the

Sample-space in to 'n'-parts.

We have A_i $\cap$ A_j $\ne$]f for i $\ne$]j, i $⩽$i, j $⩽$n

And $\sum _{i=1}^{n}P(A_i)=1$

Total Probability Theorem:

Let 'A' be any event of S. We can write A = (A₁ $\cap$ A) $\cup$ (A₂ $\cap$ A) L $\cup$(A_n $\cap$ A). As A₁, A₂ L A_n are mutually exclusive, (A₁ $\cap$ A), (A₂ $\cap$ A), L, (A_n $\cap$ A) would also be mutually exclusive.

$\Rightarrow$ P(A) = P(A₁ $\cap$ A) + P(A₂ $\cap$ A) + L + P(A_n $\cap$ A)

= P(A₁)×P(A/A₁) + P(A₂)×P(A/A₂) + L + P(A_n)×P(A/A_n)

$\Rightarrow$ P(A) = $\sum _{i=1}^{n}P(A_i)\cdot P(A/A_i)$

This is known as the total probability of the event A.

Remark:

$\bullet$ P(A/A_i) gives up the contribution of A_i in the occurrence of A.

Example -1: One bag contains four white balls and three black balls and a second bag contains three white balls and five black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from second bag is black ?

Sol: Bag – I Bag – II

4 w 3 w

3 B 5 B

Let A₁ be the event that a white ball is transferred from bag – I to bag – II and A₂ be the event that a black ball is transferred from bag – I to bag –II.

P (A₁) =$\frac{4}{7}$, P(A₂) = $\frac{3}{7}$

Let 'A' be the probability the finally a black ball is drawn from the second bag P(A/A₁) = $\frac{5}{9}$, P(A/A₂) = $\frac{6}{9}$.

Now from total probability theorem we get,

P(A) = P(A₁) . P(A/A₁) + P(A₂) . P(A/A₂) = $\frac{4}{7}.\frac{5}{9}+\frac{3}{7}.\frac{6}{9}=\frac{38}{63}$

Bayes' Theorem:

This theorem at times is also called inverse probability theorem.

Let us consider any event 'A' of sample space 'S' (as in the previous section). This event would have occurred due to the different causes (or due to the occurrence of any of the event A₁, A₂, L A_n).

Now, let us say that event A is found to have occurred and we have to find the probability that it has occurred to the occurrence of cause, say A_i.

That means we are interested in P(A_i/A). These types of problems are solved with the help of Bayes' theorem.

From total probability theorem we get P(A) = $\sum _{i=1}^{n}P(A_i)\cdot P(A/A_i)$

Also, P(A_i/A) = $\frac{P(A_i\cap A)}{P(A)}=\frac{P(A_i)\cdot P(A/A_i)}{P(A)}$

$\Rightarrow$ P(A_i/A) = $\frac{P(A_i)\cdot P(A/A_i)}{\sum _{i=1}^{n}P(A_i)\cdot P(A/A_i)}$

This result is known as Bayes' theorem.

Example -2: A bag 'A' contains 2 w and 3 red balls, a bag 'B' contains 4 w and 5 black balls. A bag is selected randomly and a ball is drawn from it. Drawn ball is observed to be white. Find the probability that bag 'B' was selected.

Sol: Bag A Bag B

2 w, 3R 4 w, 5B

Let A₁ be the event that bag 'A' is selected and A₂ be the event that bag B is selected.

P(A₁) = P(A₂) =1/2

Let 'A' be the event that a white ball is drawn from the selected bag.

$\Rightarrow$ P(A/A₁) = 2/5, P(A/A₂) = $\frac{4}{9}$

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

= $\frac{1}{2}.\frac{2}{5}+\frac{1}{2}.\frac{4}{9}=\frac{1}{2}\left(\frac{2}{5}+\frac{4}{9}\right)=\frac{38}{90}$

Finally, P(A₂/A) = $\frac{P(A_2).P(A/A_2)}{P(A)}$ = $\frac{(1/2).(4/9)}{(38/90)}=\frac{90.4}{18.38}=\frac{10}{19}$

Example -3: A card from a pack of 52 cards is lost. From the remaining cards, two cards are drawn and are found to be spades. Find the probability that missing card is also a spade.

Sol: Let A₁ be the event that missing card is spade and A₂ be event that missing card is non-spade.

$\Rightarrow$ P(A₁) = $\frac{1}{4}$, P(A₂) = $\frac{3}{4}$

Let 'A' be event that 2 spade cards are drawn from the remaining cards,

$P\left(\frac{A}{A_1}\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_2}andP\left(\frac{A}{A_2}\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}$

$\Rightarrow$ P(A) = P(A₁) ×$P\left(\frac{A}{A_1}\right)+P(A_2)\cdot P\left(\frac{A}{A_2}\right)$

= $\frac{1}{4}\frac{{}^{12}{C}_2}{{}^{51}{C}_2}+\frac{3}{4}\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{1}{4{\cdot }^{51}{C}_2}\left[{}^{12}{C}_2+3{\cdot }^{13}{C}_2\right]$

Now $P\left(\frac{A_1}{A}\right)=\frac{P(A_1)P\left(\frac{A}{A_1}\right)}{P(A)}=\frac{\frac{1}{4}\frac{{}^{12}{C}_2}{{}^{51}{C}_2}}{\frac{1}{4{\cdot }^{51}{C}_2}\left[{}^{12}{C}_2+3{\cdot }^{13}{C}_2\right]}=\frac{11}{50}$