Bayes Theorem: Statement, Proof, Formula & Examples

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Bayes Theorem is a theorem that is used to describe the occurrence of an event based on the condition that has already happened. It is a mathematical formula which is used to calculate the conditional probability of an event.

Bayes theorem is also known as Bayes law or Bayes rule.
The theorem is named after mathematician Thomas Bayes.
The probability defines the likelihood of an event.
It can represented as finding P(A | B) when P(B | A) is given.
The concept of Bayes Theorem is based on Bayesian Inference.
The most common application of Bayes theorem is in the healthcare industry.
With the help of the theorem, patient symptoms can be computed easily.
The observation is based on patient condition.

Read More: Complex Numbers and Quadratic Equations

Table of Content

What is Bayes Theorem?
Bayes Theorem Statement
Bayes Theorem Proof
Solved Examples of Bayes Theorem
Application of Bayes Theorem
Things to Remember
Sample Questions

Key Terms: Bayes Theorem, Probability, Event, Conditional Probability, Set, Hypothesis, Statistics, Variable

What is Bayes Theorem?

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Bayes Theorem is a part of conditional probability that helps in finding the probability of an event, based on prior knowledge of conditions that might be related to that event. Bayes Theorem is mathematically stated as:-

P(A|B) = P(B|A).P(A)/P(B)

where Events A and B are mutually exhaustive events.
P(A) and P(B) are the probabilities of the events A and B, respectively.
P(A|B) is the conditional probability of the occurrence of event A, given that event B has occurred.
P(B|A) is the conditional probability of the occurrence of event B, given that event A has already occurred.

Bayes Theorem

Bayes Theorem Statement

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Conisder E₁, E₂,…, E_n be a set of events of a sample S, where all the events E₁, E₂,…, E_n have nonzero probability of occurrence. Let A be any event or a condition associated with sample S, then according to Bayes theorem,

P(E_i|A) = P(E_i)P(A|E_i) / ∑P(E_k)P(A|E_k)

where, P(E_i) is the probability of event E_i.
P(E_i|A) is the probability of event E_igiven that event A has already occurred.
P(A|E_i) is the probability of event A given that event E_i has already occurred.
The summation in the denominator is to be done from i=1 to k=number of events.

This formula holds good for more events. The formula works as Bayes’ Theorem is not restricted to two events.

The video below explains this:

Bayes Theoram Detailed Video Explanation:

The pre-requisite for the Bayes' Theorem is that all the events must be mutually exhaustive. The formula will become clearer as we move forward to the examples.

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Bayes Theorem Proof

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Bayes Theorem can be proved by using total probability and conditional probability formulas.

A = A ∩ S

A ∩ (E₁,E₂,E₃,...,En)

A = (A ∩ E₁) ∪ (A ∩ E₂) ∪ (A ∩ E₃)...∪ (A ∩ E_n)

P(A) = P[(A ∩ E₁)) ∪ (A ∩ E₂) ∪ (A ∩ E₃))....∪ (A ∩ E_n)]

Since A and B are disjoint sets, then P(A∪B) = P(A) + P(B)

According to multiplication theorem,

P(A) = P(E). P(A|E₁) + P(E). P(A|E₂) + P(E). P(A|E₃)......+ P(A|E_n)

Thus total probability of P(A) = ∑n_i=1P(E_i)P(A|E_i),i=1,2,3,...,n --- (1)

Recalling the conditional probability, we get

P(E_i|A)=P(E_i∩ A) / P(A),i=1,2,3,...,n ---(2)

Using the formula for conditional probability,

P(E_i ∩ A)=P(A|E_i) P(E_i) --- (3)

Substituting equations (1) and (3) in equation (2)

P(E_i|A) = P(E_i)P(A|E_i) / ∑P(E_k)P(A|E_k)

Hence, Bayes rule is proved.

Bayes Theorem

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Solved Examples of Bayes Theorem

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Given below are some example of calculating the value of sample by using Bayes Theorem.

Example 1. A doctor is to visit a patient. From 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?

Sol. Let A be the eventy for getting late.

Let E₁, E_2, E_3, annd E₄ be the events for arriving from train, bus, scooter, and other transportations, respectively.

Therefore, P(E1) = 0.3

P(E₂) = 0.2

P(E₃) = 0.1

P(E₄ ) = 0.4 , and

P(A|E₁) = 0.25

P(A|E₂) = 0.33

P(A|E₃) = 0.083

P(A|E₄) = 0

Required probability: P(E₁|A) = P(E₁)P(A|E₁) / ∑P(E_k)P(A|E_k)

P(E₁|A) = 0.3*0.25/ [0.3*0.25 + 0.2*0.33 + 0.1*0.083 + 0.4*0]

= 0.5

Read More: Theoretical Probability

Example 2. 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?

Sol. Let E₁ , E₂ and E₃ be the events that boxes I, II and III are chosen, respectively.

Then P(E₁ ) = P(E₂ ) = P(E₃) = 1/3

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

Then, P(A|E₁ ) = 1

P(A|E₂ ) = 0

P(A|E₃ ) = 1/2

Required Probability: P(E₁|A) = P(E₁)P(A|E₁) / ∑P(E_k)P(A|E_k)

=1/3 *1/ [1/3 *1 + 1/3 *0 + 1/3 * ½ ]

= 2/3

Applications of Bayes Theorem

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The various applications of Bayes theorem are as follows:

Bayes Theorem can be used in the field of medicine to determine the accuracy of blood tests.
Meteorologists can use to determine the forecast of weather using Bayes theorem.
A number of popular puzzles, like the child problem and the third prisoner problem, can be solved using the theorem.
It can be used to detect the probability of cancer without knowing the age.

Read More: Probability distribution

Things to Remember

Bayes Theorem is used to determine the probability of an event based on a condition.
The theorem is divided into two categories, namely dependent and independent events.
Bayes Theorem formula for two variables is given as P(A|B) = P(B|A).P(A)/P(B).
Conditional probability is a type of probability when one or more events are true.
The theorem can be used by police and government for drug testing.

Read More: Sum of Squares

Sample Questions

Ques. Amit has two bags. Bag I has 8 red and 5 blue balls and bag II has 6 red and 10 blue balls. Amit draws a ball at random and it turns out to be red. Determine the probability that the ball was from the bag I. (5 marks)

Ans. Assume A to be the event of drawing a red ball. We know that the probability of choosing a bag for drawing a ball is 1/2, that is,

P(X) = P(Y) = 1/2

Let X and Y be the events that the ball is from the bag I and bag II, respectively.

P(drawing a red ball from the bag I) = P(A|X) = 8/13

Similarly, P(drawing a red ball from bag II) = P(A|Y) = 6/16

We need to determine the value of P(the ball drawn is from the bag I given that it is a red ball), that is, P(X|A). To determine this we will use Bayes Theorem. Using Bayes theorem, we have the following:

P(X|A)=P(A|X)P(X) / P(A|X)P(X)+P(A|Y)P(Y)

[((8/13)(1/2)) / (8/13)(1/2) +( 6/16)(1/2)]

= 0.62

Ques. A man is known to speak the truth 3 out of 4 times. He throws a die and reports that the number obtained is a four. Find the probability that the number obtained is actually a four. (5 marks)

Ans. Let A be the event that the man reports that number four is obtained.

Let E₁ be the event that four is obtained and E₂ be its complementary event.

Then, P(E₁) = Probability that four occurs = 1/6.

P(E₂) = Probability that four does not occur = 1- P(E₁) = 1 – (1/6) = 5/6.

Also, P(A|E₁)= Probability that man reports four and it is actually a four = 3/4

P(A|E₂) = Probability that man reports four and it is not a four = 1/4.

By using Bayes’ theorem, probability that number obtained is actually a four, P(E₁|A)

= P(X|A)=P(A|X)P(X) / P(A|X)P(X)+P(A|Y)P(Y)

= (1/6 x 3/4) / (1/6 x 3/4) + (5/6 x 1/4)

= 0.375

Ques.It is estimated that 40% of emails are spam emails. Some software has been applied to filter these spam emails before they reach your inbox. A certain brand of software claims that it can detect 99% of spam emails, and the probability for a false positive (a non-spam email detected as spam) is 5%.(4 marks)

Ans.A = event that an email is detected as spam

B = event that an email is spam
B^_c = event that an email is not spam
We know P(B) = P(B^c ) = 0.40,
P(A | B) = 0.99, P(A | Bc ) = 0.05.
Hence by the Bayes theorem we have P(B^c | A) = P(A | Bc )P(B^c ) P(A | B) P(B) + P(A|B^c )P(B^c )
0.05 × 0.4 / (0.05 × 0.4 + 0.99 × 0.6) = 0.0325

Ques.Three persons A, B and C have applied for a job in a private company. The chance of their selections is in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve the profits of the company are 0.2, 0.3 and 0.4, respectively. If the change does not take place, find the probability that it is due to the appointment of C.(5marks)

Ans.Let E₁: person A get selected

E₂: person B get selected

E₃: person C get selected

A: Changes introduced but profit not happened

Now, P(E₁) = 1/(1+2+4) = 1/7

P(E₂) = 2/7 and P(E₃) = 4/7

P(A|E₁) = P(Profit not happened by the changes introduces by A) = 1 – P(Profit happened by the changes introduces by A) = 1 – 0.2 = 0.8

P(A|E₂) = P(Profit not happened by the changes introduces by B) = 1 – P(Profit happened by the changes introduces by B) = 1 – 0.3 = 0.7

P(A|E₃) = P(Profit not happened by the changes introduces by C) = 1 – P(Profit happened by the changes introduces by C) = 1 – 0.4 = 0.6

We have to find the probability of not happening profit due to selection of C

P(E₃|A) = 0.6 x 4/7 / [ 0.6 x 4/7 + 0.8 x 1/7 + 0.7 x 2/7]

= 0.394

Ques. Riya has two bags. Bag I has 2 red and 5 blue balls and bag II has 3 red and 4 blue balls. Riya draws a ball at random and it turns out to be red. Determine the probability that the ball was from the bag I. (5 marks)

Ans. Assume A to be the event of drawing a red ball. We know that the probability of choosing a bag for drawing a ball is 1/2, that is,

P(X) = P(Y) = 1/2

Let X and Y be the events that the ball is from the bag I and bag II, respectively.

P(drawing a red ball from the bag I) = P(A|X) = 2/7

Similarly, P(drawing a red ball from bag II) = P(A|Y) = 3/7

P(X|A)=P(A|X)P(X) / P(A|X)P(X)+P(A|Y)P(Y)

[((2/7)(1/2)) / (2/7)(1/2) +( 3/7)(1/2)]

= 0.4

Ques. A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3/5 , 1/5 ,1/5 ,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.(4 marks)

Ans. Let A be the eventy for getting late.

Let E₁, E_2, E_3, annd E₄ be the events for arriving from train, bus, scooter, and other transportations, respectively.

Therefore, P(E₁) = 0.6

P(E₂) = 0.2

P(E₃) = 0.2

P(E₄ ) = 0.4 , and

P(A|E₁) = 0.25

P(A|E₂) = 0.33

P(A|E₃) = 0.083

P(A|E₄) = 0

Required probability: P(E₁|A) = P(E₁)P(A|E₁) / ∑P(E_k)P(A|E_k)

P(E₁|A) = 0.6 * 0.25/ [0.6 * 0.25 + 0.2 * 0.33 + 0.2 * 0.083 + 0.4 * 0]

= 0.644

Ques. What is the difference between conditional probability and bayes theorem. (3 marks)

Ans. The difference between conditional probability and bayes theorem are as follows:

Conditional Probability	Bayes Theorem
Conditional Probability is a probability that determines the probability of event A based on the other event B.	Bayes Theorem is based on conditional probability and occur when certain conditions are met.
Its formula is P(A\|B) = P(A∩B) / P(B)	Its formula is P(A\|B) = P(B\|A)P(A) / P(B)
The formula will given the value of A when B is already given.	The formula will given the value of A when B is already given and vice versa.

Ques. Consider that the chances of a person having a skin disease are 50%. Assuming that skin creams and drinking enough water reduces the risk of skin disease by 30% and prescription of a certain drug reduces its chance by 20%. At a time, a patient can choose any one of the two options with equal probabilities. It is given that after picking one of the options, the patient selected at random has the skin disease. Find the probability that the patient picked the option of skin screams and drinking enough water using the Bayes theorem.(5 marks)

Ans. Let E₁: The patient uses skin creams and drinks enough water; E₂: The patient uses the drug and A: The selected patient has the skin disease

P(E₁) = P(E₂) = 1/2

Using the probabilities known to us, we have

P(A|E₁) = 0.5 × (1-0.3) = 0.35

P(A|E₂) = 0.5 × (1-0.2) = 0.4

Using Bayes rule, the probability that the selected patient uses skin creams and drinks enough water is given by,

P(E₁|A) = P(A|E₁)P(E1) / [(P(A|E₁)P(E₁))+(P(A|E₂)P(E₂))]

= (0.35 × 0.5)/(0.35 × 0.5 + 0.40 × 0.5)

= 0.35/(0.35 + 0.40)

= 0.466

The probability that the patient picked the first option is 0.466

Ques. A man is known to speak the truth 2/3 times. He draws a card and reports it is king. Find the probability that it is actually a king. (4 marks)

Ans. Let E be the event that the man reports that king is drawn from the pack of cards

A be the event that the king is drawn

B be the event that the king is not drawn.

Then we have P(A) = probability that king is drawn = 1/4

P(B) = probability that king is not drawn = 3/4

P(E/A) = Probability that the man says the truth that king is drawn when actually king is drawn = P(truth) = 2/3

P(E/B)= Probability that the man lies that king is drawn when actually king is drawn = P(lie) = 1/3

Then according to Bayes formula, the probability that it is actually a king = P(A/E)

=P(A) x P(E|A) / P(A) x P(E|A) + P(B) x P(E|B)

= [1/4 × 2/3] ÷[(1/4 × 2/3) + (1/3 × 3/4)]

= 1/6 ÷5/12

= 1/6 × 12/5

=0.4

Ques. In shop A, 50 tin pure ghee and 60 tin adulterated ghee are kept for sale while in shop B, 30 tin pure ghee and 20 tin adulterated ghee are there. One tin of ghee is purchased from one of the shops randomly and it is found to be adulterated. Find the probability that it is purchased from shop B. (4 marks)

Ans. Suppose E₁ = event of choosing shop A

E₂ = event of choosing shop B

A = event of purchasing adultrated tin of ghee

P(E₁) = ½ and P(E₂) = ½

P(A|E₁) = P(purchasing adultrated ghee from shop A) = 60/110 = 6/11

P(A|E₂) = P(purchasing adultrated ghee from shop B) = 20/50 = 2/5

Therefore,

P(E₂|A) = 2/5 x ½ / [ 2/5 x ½ + 6/11 x ½]

= 1/ 5 / [ 1/5 + 3/11]

= 0.2 / 0.4727

= 0.423

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Bayes Theorem: Statement, Proof, Formula & Examples

What is Bayes Theorem?

Bayes Theorem Statement

Bayes Theoram Detailed Video Explanation:

Bayes Theorem Proof

Solved Examples of Bayes Theorem

Applications of Bayes Theorem

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

5.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

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Bayes Theorem: Statement, Proof, Formula & Examples

What is Bayes Theorem?

Bayes Theorem Statement

Bayes Theoram Detailed Video Explanation:

Bayes Theorem Proof

Solved Examples of Bayes Theorem

Applications of Bayes Theorem

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.On the basis of the above information, answer the following questions :

3.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

5. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

6. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

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1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

5.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]