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’ Theorem Notation

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:

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’ theorem Examples

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?

Answer

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.

Answer

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.

Answer

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.

Answer

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.

Answer

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%