Questions on Baye's Theorem

Bayes' Theorem

Definition, Formula and 10 University-Level Solved Problems

Definition of Bayes' Theorem

Bayes' theorem is used to calculate the probability of an event after new information or evidence has become available.

Let A and B be two events such that

P(B) > 0.

Then Bayes' theorem is

P(A | B) = [P(B | A) P(A)] / P(B)

Here:

• P(A | B) = posterior probability
• P(B | A) = likelihood
• P(A) = prior probability
• P(B) = probability of the evidence

Figure 1: Bayes' theorem updates prior probability using new evidence.

General Form

If A₁, A₂, ..., A_n form a partition of the sample space, then

P(A_i | B) = [P(A_i) P(B | A_i)] / Σ P(A_j) P(B | A_j)

Example 1: Two Machines

A factory has two machines M₁ and M₂. Machine M₁ produces 60% of the items and M₂ produces 40%. Their defective rates are 2% and 5%, respectively. An item selected randomly is found to be defective. Find the probability that it was produced by M₂.

Figure 2: Probability tree for the two machines.

Show Detailed Solution

Let D denote the event that an item is defective.

P(M₁)=0.60,   P(M₂)=0.40

P(D | M₁)=0.02

P(D | M₂)=0.05

First calculate the total probability of a defective item.

P(D) = P(M₁)P(D | M₁) + P(M₂)P(D | M₂)

= (0.60)(0.02) + (0.40)(0.05)

= 0.012+0.020 = 0.032

Now apply Bayes' theorem:

P(M₂ | D) = [P(M₂)P(D | M₂)] / P(D)

= (0.40)(0.05)/0.032

= 0.020/0.032 = 0.625

Answer: 0.625 = 62.5%

Example 2: Medical Diagnosis

A disease affects 1% of a population. A diagnostic test has 95% sensitivity and a 4% false-positive rate. If a person tests positive, find the probability that the person actually has the disease.

Show Detailed Solution

Let D denote disease and + denote a positive test.

P(D)=0.01

P(Dᶜ)=0.99

P(+ | D)=0.95

P(+ | Dᶜ)=0.04

The total probability of a positive test is

P(+) = P(+ | D)P(D) + P(+ | Dᶜ)P(Dᶜ)

= (0.95)(0.01) + (0.04)(0.99)

= 0.0095+0.0396 = 0.0491

Now apply Bayes' theorem.

P(D | +) = [P(+ | D)P(D)]/P(+)

= 0.0095/0.0491

≈ 0.1935

Answer: Approximately 19.35%

A test with high sensitivity does not necessarily imply that a positive result means a very high probability of disease. The prevalence of the disease also matters.

Example 3: Selection of an Urn

There are three urns:

U₁: 2 red and 3 blue balls
U₂: 4 red and 1 blue ball
U₃: 3 red and 3 blue balls

An urn is selected randomly and one ball is drawn. If the ball is red, find the probability that it came from U₂.

Show Detailed Solution

Since the urn is selected randomly,

P(U₁)=P(U₂)=P(U₃)=1/3

The conditional probabilities of drawing red are

P(R | U₁)=2/5

P(R | U₂)=4/5

P(R | U₃)=3/6=1/2

Using the total probability theorem,

P(R) = (1/3)(2/5) + (1/3)(4/5) + (1/3)(1/2)

= 17/30

By Bayes' theorem,

P(U₂ | R) = [(1/3)(4/5)]/(17/30)

= 8/17

Answer: 8/17 ≈ 0.4706

Example 4: University Examination

At a university, 55% of students are undergraduate and 45% are postgraduate. Among undergraduate students, 70% pass an examination, whereas 85% of postgraduate students pass. If a randomly selected student has passed, find the probability that the student is postgraduate.

Show Detailed Solution

Let U denote undergraduate, G postgraduate, and P passing.

P(U)=0.55

P(G)=0.45

P(P | U)=0.70

P(P | G)=0.85

The total probability of passing is

P(P) = (0.55)(0.70) + (0.45)(0.85)

= 0.385+0.3825 = 0.7675

Now,

P(G | P) = [(0.45)(0.85)]/0.7675

= 0.3825/0.7675

≈0.4984

Answer: Approximately 49.84%

Example 5: Three Production Plants

Plants A, B and C produce 30%, 45% and 25% of a company's total products. Their defective rates are 1%, 2% and 4%, respectively. A product is found to be defective. Find the probability that it came from plant C.

Show Detailed Solution

P(A)=0.30, P(B)=0.45, P(C)=0.25

P(D | A)=0.01

P(D | B)=0.02

P(D | C)=0.04

Total probability of defect:

P(D) = (0.30)(0.01) + (0.45)(0.02) + (0.25)(0.04)

= 0.003+0.009+0.010 = 0.022

By Bayes' theorem,

P(C | D) = [(0.25)(0.04)]/0.022

= 0.010/0.022 = 5/11

Answer: 5/11 ≈ 45.45%

Example 6: Spam Email Detection

Suppose 20% of incoming emails are spam. A particular word occurs in 70% of spam emails but only 10% of non-spam emails. If an email contains the word, find the probability that it is spam.

Show Detailed Solution

Let S denote spam and W denote occurrence of the word.

P(S)=0.20

P(Sᶜ)=0.80

P(W | S)=0.70

P(W | Sᶜ)=0.10

Total probability that the word appears:

P(W) = (0.70)(0.20) + (0.10)(0.80)

= 0.14+0.08 = 0.22

Therefore,

P(S | W) = 0.14/0.22 = 7/11

Answer: 7/11 ≈ 63.64%

Example 7: Insurance Risk

An insurance company classifies 30% of its drivers as high-risk and 70% as low-risk. The annual accident probabilities are 20% and 5%, respectively. If a driver has an accident, find the probability that the driver belongs to the high-risk group.

Show Detailed Solution

P(H)=0.30

P(L)=0.70

P(A | H)=0.20

P(A | L)=0.05

Total probability of accident:

P(A) = (0.30)(0.20) + (0.70)(0.05)

= 0.060+0.035 = 0.095

Then

P(H | A) = 0.060/0.095

= 12/19 ≈0.6316

Answer: Approximately 63.16%

Example 8: Selection of a Biased Coin

A box contains three coins. Coin C₁ is fair, coin C₂ shows heads with probability 0.75, and coin C₃ shows heads with probability 0.90. A coin is selected randomly and tossed once. If the result is a head, find the probability that C₃ was selected.

Show Detailed Solution

P(C₁)=P(C₂)=P(C₃)=1/3

P(H | C₁)=0.50

P(H | C₂)=0.75

P(H | C₃)=0.90

Total probability of obtaining a head:

P(H) = (1/3)(0.50+0.75+0.90)

= 2.15/3

Then

P(C₃ | H) = [(1/3)(0.90)]/(2.15/3)

= 0.90/2.15 ≈0.4186

Answer: Approximately 41.86%

Example 9: Reliability of Sensors

Type A sensors constitute 70% of a monitoring system and type B sensors constitute 30%. Type A gives a false alarm with probability 0.03, while type B gives a false alarm with probability 0.08. A false alarm occurs. Find the probability that it came from a type B sensor.

Show Detailed Solution

P(A)=0.70

P(B)=0.30

P(F | A)=0.03

P(F | B)=0.08

Total probability of false alarm:

P(F) = (0.70)(0.03) + (0.30)(0.08)

= 0.021+0.024 = 0.045

Using Bayes' theorem,

P(B | F) = 0.024/0.045

= 8/15 ≈0.5333

Answer: Approximately 53.33%

Example 10: Diagnostic Test with Sensitivity and Specificity

A disease affects 5% of a population. A test has 90% sensitivity and 95% specificity. A person receives a positive result. Find the probability that the person actually has the disease.

Show Detailed Solution

The sensitivity is

P(+ | D)=0.90

The specificity is

P(- | Dᶜ)=0.95

Therefore the false-positive probability is

P(+ | Dᶜ)=1-0.95=0.05

Also,

P(D)=0.05

P(Dᶜ)=0.95

Total probability of a positive result:

P(+) = (0.90)(0.05) + (0.05)(0.95)

= 0.045+0.0475 = 0.0925

Using Bayes' theorem,

P(D | +) = [(0.90)(0.05)]/0.0925

= 0.045/0.0925 ≈0.4865

Answer: Approximately 48.65%

Although the test has 90% sensitivity, this does not mean that a positive test gives a 90% probability of disease. The prior probability of disease and the false-positive rate must also be considered.

Important Bayes' Theorem Formulas

1. Conditional Probability

P(A | B) = P(A ∩ B) / P(B)

2. Bayes' Theorem

P(A | B) = [P(B | A)P(A)] / P(B)

3. Total Probability Theorem

P(B) = Σ P(B | A_i)P(A_i)

4. General Bayes' Formula

P(A_i | B) = [P(B | A_i)P(A_i] / Σ[P(B | A_j)P(A_j)]

5. Bayesian Interpretation

Posterior ∝ Likelihood × Prior

Bayes' Theorem — University Level Probability

Try each problem independently and then click Show Detailed Solution to verify your answer.