# Bayes’ Theorem: Definition, Notation, Examples

Thomas Bayes named a theory about conditional probability. He was British mathematician in 18th century. The theorem is Bayes’ Theorem.

Bayes’ theorem is used in statistics, medicine and pharmacology, finance, etc.

## Definition of Bayes’ Theorem

Bayes’ theorem is a probability theorem when we have known another probability before. If the probability of blue ball outcomes is ¼ then Bayes’ theorem is used to determine the probability of red ball outcomes after the blue ones.

In another words, Bayes’ theorem is a way to predict existing probability after the previous condition.

Bayes’ theorem is also called as Bayes’ rule.

## Bayes’ TheoremNotation

If there is two occurrences,

A : blue ball outcome

B : red ball outcome

Then probability of blue ball outcome happened given that red ball happened is

P(A|B) = {P(A) P(B|A)} / P(B)

It is called as Bayes’ Theorem

Where,

• P(A) : probability of blue ball outcome
• P(B) : probability of red ball outcome
• P(A|B) probability of blue ball outcome happens given that red ball outcome happens
• P(B|A) : probability of red ball outcome happens given that blue ball outcome happens

Based on general form of the Bayes’ theorem, there is also another form of Bayes’ theorem.

P(B) P(A|B) = P(A) P(B|A)

But, sometimes there are another form of Bayes’ Theorem.

P(B) can be describe to be longer form in some cases.

Based on the example, P(B) = {P(A) P(B|A) + P(~A) P(B|~A)}

Then the Bayes’ theorem

P(A|B) = {P(A) P(B|A)} / {P(A) P(B|A) + P(~A) P(B|~A)}

In another words,

Let A1, A2, A3, … Ai are a set of events in space S, and they have non-zero occurrence probability. Let B be any event associated with S, then Bayes’ Theorem:

Another case that using longer form of P(B):

If there are two baskets of mangoes, A and B. It has been known percentage of ready to eat mangoes in each basket. A customer takes randomly a ready to eat mango and he take from first basket. The question is determining the probability.

The solution is:

• Making a notation to “taking a ready to eat mango, mangoes in first and second basket, and ready to eat mangoes in each basket”.
M = taking a ready to eat mango
A = mangoes in first basket → P(A)
B = mangoes in second basket → P(B)
P(M|A) = ready to eat mangoes in first basket
P(M|B) = ready to eat mangoes in second basket
• Then the Bayes’ theorem is

P(A|M) = {P(A)P(M|A)} / {P(A)P(M|A) +P(B)P(M|B)}

Note: look at examples no.3 for complete case.

## Bayes’ theoremExamples

1. In a class there are 50% of the boys like mathematics and 45% of the boys like physics.
Given that 20% of those that like mathematics and physics, what percent of those that like physics and mathematics?

Let

A = boys like mathematics → P(A) = 0.5

B = boys like physics → P(B) = 0.45

P(B|A) = 0.2

Then

P(A|B) = {P(A) P(B|A)} / P(B)

= (0.5)(0.2) / (0.45)

= 0.222

= 22%

Therefore, percentage of those that like physics and mathematics is 22%

2. In an office there are two favorite drink, they are coffee and tea. 65% of employees like tea and 30% of them like to drink coffee. But, 40% of them like coffee and tea. Determine percentage of employees that prefer to drink tea and coffee.

Let

M = employee that like to drink coffee → P(M) = 0.3

N = employee that like to drink tea → P(N) = 0.65

P(N|M) = 0.4

Then

P(M|N) = {P(M) P(N|M)} / P(N)

= (0.3)(0.4) / (0.65)

= 0.185

= 18.5%

Therefore, percentage of employees that prefer to drink tea and coffee is 18.5%.

3. If there are two baskets of mangoes, A and B. 70% mangoes in basket A is ready to eat and 25% mangoes in basket B is not ready to eat yet. A customer takes randomly a ready to eat mango and he take from first basket. Determine the probability.

Let,

M = taking a ready to eat mango

A = mangoes in first basket → P(A) = 50% = 0.5

B = mangoes in second basket → P(B) = 50% = 0.5

P(M|A) = ready to eat mangoes in first basket = 70% = 0.7

P(M|B) = ready to eat mangoes in second basket = 100% – 25% = 75% = 0.75

Then,

P(A|M) = {P(A)P(M|A)} / {P(A)P(M|A) +P(B)P(M|B)}

= {(0.5)(0.7)} / {(0.5)(0.7) + (0.5)(0.75)}

= (0.35) / (0.35+0.375)

= 0.35 / 0.725

= 0.4827

= 48.27%

4. Arnold has two pockets. The first has 7 chocolate and 2 strawberry candies and the second has 5 chocolate and 9 strawberry candies. He takes a candy at random and it turns out to be chocolate. Determine the probability that the chocolate candy was from the first pocket.

First pocket: 7 chocolate and 2 strawberry, total 9 candies

Second pocket: 5 chocolate and 9 strawberry, total 14 candies

Let

A = event from first pocket

B = event from second pocket

Then probability of candy from first or second pocket is

P(A) = P(B) = ½ = 0.5 (because there are only two pocket)

Next,

X = event taking a chocolate candy

Probability of chocolate candy from first pocket is P(X|A) = 7/9 = 0.78

Probability of chocolate candy from second pocket is P(X|B) = 5/14 = 0.36

It will determine chocolate candy came from first pocket.

P(the taking candies is from first pocket given that it is chocolate candy)

P(A|X) = {P(A).P(X|A)} / {P(A).P(X|A) + P(B).P(X|B)}

= {(0.5)(0.78)} / {(0.5)(0.78) + (0.5)(0.36)

= (0.39) / (0.39 + 0.18)

= 0.39 / 0.57

= 0.68

Therefore, the probability that the chocolate candy was from the first pocket is 0.68.

5. In a restaurant there is a drug test for waiters dan chef that is 98% accurate, meaning 98% of the time it shows a true positive result for someone using the drug and 98% of the time it shows a true negative result for non-users of the drug.

Assume 0.8% of people use the drug. If a person selected at random tests positive for the drug, determine the probability the person is actually a user of the drug.

Let

A = users of the drug → P(A) = 0.008

B = non-users of the drug → P(B) = 1 – 0.001 = 0.992

M = people with positive result

P(M|A) = 0.98

P(M|B) = 1 – 0.98 = 0.02

Then,

P(A|M) = {P(M|A).P(A)} / {P(M|A).P(A) + P(M|B).P(B)}

= {(0.98)(0.008)} / {(0.98)(0.008) + (0.02)(0.992)}

= 0.0078 / (0.0078 + 0.01984)

= 0.0078 / 0.02764

= 0.282

= 28,2%