`color{red} ♦` Bayes' Theorem

`color{red} ♦` Partition of a sample space

`color{red} ♦` Theorem of total probability

`color{red} ♦` Prove the Bayes' theorem.

`color{red} ♦` Partition of a sample space

`color{red} ♦` Theorem of total probability

`color{red} ♦` Prove the Bayes' theorem.

`=>` Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags. We can find the probability of selecting any of the bags (i.e.`1/2` ) or probability of drawing a ball of a particular colour (say white) from a particular bag (say Bag I).

`=>` A set of events `E_1, E_2, ..., E_n` is said to represent a partition of the sample space S if

(a) `E_ i ∩ E_j = φ, i ≠ j, i, j = 1, 2, 3, ..., n`

(b) `E_1 ∪ Ε_2 ∪ ... ∪ E_n= S` and

(c) `P(E_i) > 0` for all i = 1, 2, ..., n.

`=>` In other words, the events `E_1, E_2, ..., E_n` represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.

`=>` As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy` E ∩ E′ = φ` and` E ∪ E′ = S`.

`=>` From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any two events associated with a sample space S, then the set` {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}` is a partition of the sample space S. It may be mentioned that the partition of a sample space is not unique. There can be several partitions of the same sample space.

(a) `E_ i ∩ E_j = φ, i ≠ j, i, j = 1, 2, 3, ..., n`

(b) `E_1 ∪ Ε_2 ∪ ... ∪ E_n= S` and

(c) `P(E_i) > 0` for all i = 1, 2, ..., n.

`=>` In other words, the events `E_1, E_2, ..., E_n` represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.

`=>` As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy` E ∩ E′ = φ` and` E ∪ E′ = S`.

`=>` From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any two events associated with a sample space S, then the set` {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′}` is a partition of the sample space S. It may be mentioned that the partition of a sample space is not unique. There can be several partitions of the same sample space.

`=>` Let `{E_1, E_2,...,E_n}` be a partition of the sample space S, and suppose that each of the events `E_1, E_2,..., E_n` has nonzero probability of occurrence. Let A be any event associated with S, then

`color{blue}{P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + ... + P(E_n) P(A|E_n)}`

`color{blue}{= sum_(j=1)^n P(E_j) P(A|E_j)}`

`color{red} "Proof"` Given that `E_1, E_2,..., E_n` is a partition of the sample space S (Fig). Therefore,

`color{orange} {S = E_1 ∪ E_2 ∪ ... ∪ E_n}` ... (1)

and `E_i ∩ E_j = φ, i ≠ j, i, j = 1, 2, ..., n`

Now, we know that for any event A,

`A = A ∩ S`

`= A ∩ (E_1 ∪ E_2 ∪ ... ∪ E_n)`

`= (A ∩ E_1) ∪ (A ∩ E_2) ∪ ...∪ (A ∩ E_n)`

Also `A ∩ E_i` and `A ∩ E_j` are respectively the subsets of `E_i` and `E_j` . We know that

`E_i` and `E_j` are disjoint, for i ≠ j , therefore, `A ∩ E_i` and `A ∩ E_j` are also disjoint for all

`i ≠ j, i, j = 1, 2, ..., n`.

Thus, `P(A) = P [(A ∩ E_1) ∪ (A ∩ E_2)∪ .....∪ (A ∩ E_n)]`

`= P (A ∩ E_1) + P (A ∩ E_2) + ... + P (A ∩ E_n)`

Now, by multiplication rule of probability, we have

`P(A ∩ E_i) = P(E_i) P(A|E_i)` as `P (E_i) ≠ 0 ∀ i = 1,2,..., n`

Therefore, `P (A) = P (E_1) P (A|E_1) + P (E_2) P (A|E_2) + ... + P (E_n)P(A|E_n)`

or `color{green}{P(A) = sum_(j=1)^n P(E_j) P(A |E_j)}`

`color{blue}{P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + ... + P(E_n) P(A|E_n)}`

`color{blue}{= sum_(j=1)^n P(E_j) P(A|E_j)}`

`color{red} "Proof"` Given that `E_1, E_2,..., E_n` is a partition of the sample space S (Fig). Therefore,

`color{orange} {S = E_1 ∪ E_2 ∪ ... ∪ E_n}` ... (1)

and `E_i ∩ E_j = φ, i ≠ j, i, j = 1, 2, ..., n`

Now, we know that for any event A,

`A = A ∩ S`

`= A ∩ (E_1 ∪ E_2 ∪ ... ∪ E_n)`

`= (A ∩ E_1) ∪ (A ∩ E_2) ∪ ...∪ (A ∩ E_n)`

Also `A ∩ E_i` and `A ∩ E_j` are respectively the subsets of `E_i` and `E_j` . We know that

`E_i` and `E_j` are disjoint, for i ≠ j , therefore, `A ∩ E_i` and `A ∩ E_j` are also disjoint for all

`i ≠ j, i, j = 1, 2, ..., n`.

Thus, `P(A) = P [(A ∩ E_1) ∪ (A ∩ E_2)∪ .....∪ (A ∩ E_n)]`

`= P (A ∩ E_1) + P (A ∩ E_2) + ... + P (A ∩ E_n)`

Now, by multiplication rule of probability, we have

`P(A ∩ E_i) = P(E_i) P(A|E_i)` as `P (E_i) ≠ 0 ∀ i = 1,2,..., n`

Therefore, `P (A) = P (E_1) P (A|E_1) + P (E_2) P (A|E_2) + ... + P (E_n)P(A|E_n)`

or `color{green}{P(A) = sum_(j=1)^n P(E_j) P(A |E_j)}`

`=>` Bayes’ Theorem If `E_1, E_2 ,..., E_n` are n non empty events which constitute a partitionof sample space S, i.e. `E_1, E_2 ,..., E_n` are pairwise disjoint and `E_1∪ E_2∪ ... ∪ E_n = S` and A is any event of nonzero probability, then

`"Proof : "` `color{blue}{P(E_i | A) = (P(E_i) P(A|E_i))/(Sigma_(j =1)^n P(E_j ) P (A|E_j))}` for any i = 1, 2, 3, ..., n

`color {red} "Proof"` By formula of conditional probability, we know that

`P(E_r|A) = (P(AnnE_i))/(P(A))`

`= (P(E_i)P(A|E_i))/(P(A) )` (by multiplication rule of probability)

`= (P(E_i ) (PA | E_i))/(Sigma_(j=1)^n P(E_j) P(A| E_j))` (by the result of theorem of total probability)

`color{blue} ul"Remark"` :

The following terminology is generally used when Bayes' theorem is applied.

The events `E_1, E_2, ..., E_n` are called hypotheses.

`=>` The probability `P(E_i)` is called the priori probability of the hypothesis `E_i`

`=>` The conditional probability `P(E_i |A) `is called a posteriori probability of the hypothesis `E_i`.

`=>` Bayes' theorem is also called the formula for the probability of "causes". Since the `E_i's` are a partition of the sample space S, one and only one of the events `E_i` occurs (i.e. one of the events `E_i` must occur and only one can occur). Hence, the above formula gives us the probability of a particular `E_i` (i.e. a "Cause"), given that the event A has occurred.

The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples.

`"Proof : "` `color{blue}{P(E_i | A) = (P(E_i) P(A|E_i))/(Sigma_(j =1)^n P(E_j ) P (A|E_j))}` for any i = 1, 2, 3, ..., n

`color {red} "Proof"` By formula of conditional probability, we know that

`P(E_r|A) = (P(AnnE_i))/(P(A))`

`= (P(E_i)P(A|E_i))/(P(A) )` (by multiplication rule of probability)

`= (P(E_i ) (PA | E_i))/(Sigma_(j=1)^n P(E_j) P(A| E_j))` (by the result of theorem of total probability)

`color{blue} ul"Remark"` :

The following terminology is generally used when Bayes' theorem is applied.

The events `E_1, E_2, ..., E_n` are called hypotheses.

`=>` The probability `P(E_i)` is called the priori probability of the hypothesis `E_i`

`=>` The conditional probability `P(E_i |A) `is called a posteriori probability of the hypothesis `E_i`.

`=>` Bayes' theorem is also called the formula for the probability of "causes". Since the `E_i's` are a partition of the sample space S, one and only one of the events `E_i` occurs (i.e. one of the events `E_i` must occur and only one can occur). Hence, the above formula gives us the probability of a particular `E_i` (i.e. a "Cause"), given that the event A has occurred.

The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples.

Q 3187591487

Bag I contains 3 red and 4 black balls while another Bag II contains 5 red

and 6 black balls. One ball is drawn at random from one of the bags and it is found to

be red. Find the probability that it was drawn from Bag II.

Class 12 Chapter 13 Example 16

and 6 black balls. One ball is drawn at random from one of the bags and it is found to

be red. Find the probability that it was drawn from Bag II.

Class 12 Chapter 13 Example 16

Let `E_1` be the event of choosing the bag `I, E_2` the event of choosing the bag II

and A be the event of drawing a red ball.

Then `P(E_1) = P(E_2) = 1/2`

Also` P(A|E_1) =` P(drawing a red ball from Bag I) =3/7

and `P(A|E_2) =` P(drawing a red ball from Bag II) = 5/11

Now, the probability of drawing a ball from Bag II, being given that it is red,

is `P(E_2|A)`

By using Bayes' theorem, we have

`P(E_2/A) = ( P(E_2) P(A| E_2) )/(P(E_1) P( A | E_1) + P ( E_2) p (A| E_2)) = (1/2 xx 5/11)/( 1/2 xx 3/7 + 1/2 xx 5/11 ) = 35/68`

Q 3117591489

Given three identical boxes I, II and III, each containing two coins. In

box I, both coins are gold coins, in box II, both are silver coins and in the box III, there

is one gold and one silver coin. A person chooses a box at random and takes out a coin.

If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Class 12 Chapter 13 Example 17

box I, both coins are gold coins, in box II, both are silver coins and in the box III, there

is one gold and one silver coin. A person chooses a box at random and takes out a coin.

If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Class 12 Chapter 13 Example 17

Let `E_1, E_2` and `E_3` be the events that boxes I, II and III are chosen, respectively .

Then `P(E_1) = P(E_2) = P(E_3) = 1/3`

Also, let A be the event that ‘the coin drawn is of gold’

Then `P(A|E_1) = P`(a gold coin from bag I)` = 2/2 = 1`

`P(A|E_2) = P`(a gold coin from bag II) = 0

`P(A|E_3) = P`(a gold coin from bag III) = 1/2

Now, the probability that the other coin in the box is of gold

= the probability that gold coin is drawn from the box I.

=` P(E_1|A)`

By Bayes' theorem, we know that

`P(E_1|A) = (P(E_1) P( A| E_1))/( P(E_1 )P(A|E_1 )+P(E_2 )P(A|E_2 )+P(E_3 )P(A|E_3 ) )`

`= (1/3 xx 1)/( 1/3 xx 1 + 1/3 xx 0 + 1/3 xx 1/2 ) = 2/3 `

Q 3147691583

Suppose that the reliability of a HIV test is specified as follows:

Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of

people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as

showing HIV+ive. From a large population of which only 0.1% have HIV, one person

is selected at random, given the HIV test, and the pathologist reports him/her as

HIV+ive. What is the probability that the person actually has HIV?

Class 12 Chapter 13 Example 18

Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of

people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as

showing HIV+ive. From a large population of which only 0.1% have HIV, one person

is selected at random, given the HIV test, and the pathologist reports him/her as

HIV+ive. What is the probability that the person actually has HIV?

Class 12 Chapter 13 Example 18

Let E denote the event that the person selected is actually having HIV and A

the event that the person's HIV test is diagnosed as +ive. We need to find P(E|A).

Also E′ denotes the event that the person selected is actually not having HIV.

Clearly, {E, E′} is a partition of the sample space of all people in the population.

We are given that

`P(E) = 0.1 % = (0.1)/(100) = 0.001`

`P(E′) = 1 – P(E) = 0.999`

P(A|E) = P(Person tested as HIV+ive given that he/she

is actually having HIV)

`= 90% = 90/100 = 0.9`

and P(A|E′) = P(Person tested as HIV +ive given that he/she

is actually not having HIV)

` = 1%= 1/100 = 0.01`

Now, by Bayes' theorem

`P(E|A) = (P(E) P(A| E ))/(P(E) P(A|E) + P (E') P(A|E') )`

`= (0.001 xx 0.9)/(0.001 xx 0.9 + 0.999 xx 0.01 ) = 90/1089`

= 0.083 approx.

Thus, the probability that a person selected at random is actually having HIV

given that he/she is tested HIV+ive is 0.083.

Q 3157691584

In a factory which manufactures bolts, machines A, B and C manufacture

respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are

respectively defective bolts. A bolt is drawn at random from the product and is found

to be defective. What is the probability that it is manufactured by the machine B?

Class 12 Chapter 13 Example 19

respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are

respectively defective bolts. A bolt is drawn at random from the product and is found

to be defective. What is the probability that it is manufactured by the machine B?

Class 12 Chapter 13 Example 19

Let events `B_1, B_2, B_3` be the following :

`B_1 :` the bolt is manufactured by machine A

`B_2 :` the bolt is manufactured by machine B

`B_3 :` the bolt is manufactured by machine C

Clearly, `B_1, B_2, B_3` are mutually exclusive and exhaustive events and hence, they

represent a partition of the sample space.

Let the event E be ‘the bolt is defective’.

The event E occurs with `B_1` or with `B_2` or with `B_3`. Given that,

`P(B_1) = 25% = 0.25, P (B_2) = 0.35` and `P(B_3) = 0.40`

Again `P(E|B_1) = `Probability that the bolt drawn is defective given that it is manufactured

by machine A = 5% = 0.05

Similarly, `P(E|B_2) = 0.04, P(E|B_3) = 0.02`.

Hence, by Bayes' Theorem, we have

`P(B_2 |E) = ( P(B_2 )P (E | B_2 ) )/( P(B_1 ) P ( E | B_1 )+ P (B_2 ) P (E |B_2 )+P(B_3 )P(E|B_3 ) )`

` = (0.35 xx 0.04)/( 0.25 xx 0.05 +0.35 xx 0.40 xx 0.02)`

`= (0.0140)/(0.0345) = 28/69`

Q 3167691585

A doctor is to visit a patient. From the past experience, it is known that

the probabilities that he will come by train, bus, scooter or by other means of transport

are respectively 3/10 , 1/5 ,1/10 and 2/5 . The probabilities that he will be late are 1/4 ,1/3 , and 1/12 ,

if he comes by train, bus and scooter respectively, but if he comes by other means of

transport, then he will not be late. When he arrives, he is late. What is the probability

that he comes by train?

Class 12 Chapter 13 Example 20

the probabilities that he will come by train, bus, scooter or by other means of transport

are respectively 3/10 , 1/5 ,1/10 and 2/5 . The probabilities that he will be late are 1/4 ,1/3 , and 1/12 ,

if he comes by train, bus and scooter respectively, but if he comes by other means of

transport, then he will not be late. When he arrives, he is late. What is the probability

that he comes by train?

Class 12 Chapter 13 Example 20

Let E be the event that the doctor visits the patient late and let `T_1, T_2, T_3, T_4` be the events that the doctor comes by train, bus, scooter, and other means of transport

respectively.

Then `P(T_1) = 3/10 , P (T_2) =1/5, P(T_3) = 1/10` and `P(T_4) = 2/5` (given)

`P(E|T_1) = `Probability that the doctor arriving late comes by train = 1/4

Similarly,` P(E|T_2) = 1//3 , P(E|T_3) = 1/12` and ` P(E|T_4) = 0`, since he is not late if he

comes by other means of transport.

Therefore, by Bayes' Theorem, we have

`P(T_1|E) =` Probability that the doctor arriving late comes by train

`= (P(T_1 )P(E|T_1 ))/( P(T_1 )P( E | T_1 )+P(T_2 )P( E | T_2 )+P(T_3 )P(E|T_3 )+P(T_4 )P(E|T_4 ) )`

`= (3/10 xx 1/4 )/( 3/10 xx 1/4 + 1/5 xx 1/3 + 1/10 xx 1/12 +2/5 xx 0) = 3/40 xx 120/18 = 1/2`

Hence, the required probability is 1/2

Q 3177691586

A man is known to speak truth 3 out of 4 times. He throws a die and

reports that it is a six. Find the probability that it is actually a six.

Class 12 Chapter 13 Example 21

reports that it is a six. Find the probability that it is actually a six.

Class 12 Chapter 13 Example 21

Let E be the event that the man reports that six occurs in the throwing of the

die and let `S_1` be the event that six occurs and `S_2` be the event that six does not occur.

Then `P(S_1) =` Probability that six occurs =1/6

`P(S_2) =` Probability that six does not occur = 5/6

`P( E |S_1) = `Probability that the man reports that six occurs when six has

actually occurred on the die

= Probability that the man speaks the truth = 3/4

`P(E|S_2) =` Probability that the man reports that six occurs when six has

not actually occurred on the die

= Probability that the man does not speak the truth = 1-3/4 = 1/4

Thus, by Bayes' theorem, we get

`P(S_1|E) =` Probability that the report of the man that six has occurred is

actually a six

`= ( P(S_1 )P(E |S_1 ) )/( P(S_1 )P(E|S_1 )+P(S_2 )P(E|S_2 ) )`

`= (1/6 xx 3/4 )/( 1/6 xx 3/4 + 5/6 xx 1/4 ) = 1/8 xx 24/8 = 3/8`

Hence, the required probability is 3/8