Bayes Theorem Questions

Bayes’ Theorem is a fundamental concept in probability theory and statistics that plays a crucial role in decision-making under uncertainty. It allows us to update the probability of an event based on new evidence, helping in fields ranging from medical diagnosis to finance and artificial intelligence.

1.0What is Bayes' Theorem?

Bayes' Theorem describes how to compute the probability of a hypothesis given some evidence. It helps in understanding the relationship between prior beliefs (initial assumptions) and new evidence to compute a more accurate probability (posterior probability).

The formula for Bayes' Theorem is:

$P\left(E_i\mid E\right)=\frac{P\left(E\mid E_i\right)\cdot P\left(E_i\right)}{P(E)}$

Where:

• P(E_i |E) is the posterior probability (the probability of event E_i given that E is true).
• P(E | E_i) is the likelihood (the probability of event E given that E_i is true).
• P(E_i) is the prior probability (the initial probability of event E_i).
• P(E) is the marginal likelihood (the total probability of event E).

2.0Bayes’ Theorem: State and Prove

Statement

Bayes’ Theorem relates the conditional probability of two events and is stated as follows:

$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$

Proof:

From the definition of conditional probability:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

$P(B\mid A)=\frac{P(A\cap B)}{P(A)}$

Rearranging the second equation gives:

$P(A\cap B)=P(B\mid A)\cdot P(A)$

Substituting this into the first equation, we get:

$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$

This is the Bayes’ Theorem formula.

3.0Bayes Theorem Solved Questions

Example 1: There are three identical boxes, labeled I, II, and III, each containing two coins. Box I holds two gold coins, Box II holds two silver coins, and Box III contains one gold and one silver coin. A person randomly selects a box and draws a coin, which turns out to be gold. What is the probability that the other coin in the selected box is also gold?

Solution:

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

Then $P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

Also, Let A represent the event that "a gold coin is drawn."

Then

P(A|E₁) = P (drawing a gold coin from box I) =  $\frac{2}{2}$ = 1

P(A|E₂) = P (drawing a gold coin from box II) = 0

P(A|E₃) = P (drawing a gold coin from box III) = $\frac{1}{2}$

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

= the probability of drawing a gold coin from box I.

= P(E₁ |A)

By Bayes theorem, we know that

$P\left(E_1\mid A\right)=\frac{P\left(E_1\right)P\left(A\mid E_1\right)}{P\left(E_1\right)P\left(A\mid E_1\right)+P\left(E_2\right)P\left(A\mid E_2\right)+P\left(E_3\right)P\left(A\mid E_3\right)}$

$=\frac{\left(\frac{1}{3}\times 1\right)}{\left(\frac{1}{3}\times 1\right)+\left(\frac{1}{3}\times 0\right)+\left(\frac{1}{3}\times \frac{1}{2}\right)}=\frac{2}{3}$

Hence, the required probability is $\frac{2}{3}$.

Example 2: In a factory that produces bolts, three machines—A, B, and C—are responsible for manufacturing 25%, 35%, and 40% of the total bolts, respectively. The defect rates of these machines differ: machine A produces 5% defective bolts, machine B produces 4% defective bolts, and machine C produces 2% defective bolts. A bolt is randomly selected from the overall production, and it is found to be defective. What is the probability/likelihood that this defective bolt was produced by machine B?

Solution: 

Let events B₁, B₂, and B₃ represent the following: 

B₁: the bolt is produced by machine A

B₂: the bolt is produced by machine B

B₃: the bolt is produced by machine C

Clearly, the events B₁, B₂, and B₃ are mutually exclusive and collectively exhaustive, meaning they form a complete partition of the sample space.

Let event E represent 'the bolt is defective'.

The event E occurs along with B₁, B₂ or B₃. 

Given:

P(B₁) = 25% = 0.25, 

P(B₂) = 0.35 and 

P(B₃) = 0.40

Again P(E|B₁) represents the probability that the bolt is defective, given that it was manufactured by machine A = 5% = 0.05.

In the same way, P(E|B₂) = 0.04, 

 P(E|B₃) = 0.02.

Hence, by Bayes Theorem, we have 

$P\left(B_2\mid E\right)=\frac{P\left(B_2\right)P\left(E\mid B_2\right)}{P\left(B_1\right)P\left(E\mid B_1\right)+P\left(B_2\right)P\left(E\mid B_2\right)+P\left(B_3\right)P\left(E\mid B_3\right)}$

$=\frac{0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.40\times 0.02}$

$=\frac{0.0140}{0.0345}$

$=\frac{28}{69}$

Hence, the required probability is $\frac{28}{69}.$

Example 3 : In a test an examinee either guesses, copies or knows the answer to a multiple-choice question with four choices. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$ . Given that the probability of his answer being correct when he copied it is $\frac{1}{8}$, what is the probability that he actually knew the answer to the question, considering the probability that his answer is correct if he copied it? 

Solution: 

Let A₁ be the event that the examinee guesses that answer; A₂ the event that he copies the answer and A₃ the event that he knows the answer. Also let A be the event that he answers correctly. Then as given, we have

$\mathrm{P}\left({\mathrm{A}}_1\right)=\frac{1}{3}$,

$\mathrm{P}\left({\mathrm{A}}_2\right)=\frac{1}{6}$ ,

$\mathrm{P}\left({\mathrm{A}}_3\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}$

[We have assumed here that the events A₁, A₂ and A₃ are mutually exclusive and totally exhaustive.]

Now P(A/A1) = $\frac{1}{4}$ 

$\mathrm{P}\left(\mathrm{A}/{\mathrm{A}}_2\right)=\frac{1}{8}(asgiven)$

gain it is reasonable to take the probability of answering correctly given that he knows the answer as 1, that is, 

P(A/A₃) = 1

We have to find P(A₃/A).

By Bayes theorem, we have

$\mathrm{P}\left({\mathrm{A}}_3/\mathrm{A}\right)=\frac{P\left(A_3\right)P\left(A/A_3\right)}{P\left(A_1\right)P\left(A/A_1\right)+P\left(A_2\right)P\left(A/A_2\right)+\left(A_3\right)P\left(A/A_3\right)}$

$=\frac{\left(\frac{1}{2}\right)\times 1}{\left(\frac{1}{3}\times \frac{1}{4}\right)+\left(\frac{1}{6}\times \frac{1}{8}\right)+\left(\frac{1}{2}\right)\times (1)}$

$=\frac{24}{29}$

Hence, the required probability is $\frac{24}{29}$.

Example 4: A newly constructed bridge may fall down either due to wrong designing or by inferior material used in construction. The chance that the design is faulty is 10% and the probability of its collapse if the design is faulty is 95% and that due to bad material it is 45%. If the bridge collapses. Find the chance that it was due to wrong designing.

Solution : 

Let E₁ be the event due to wrong designing and E₂ be the event due to interior material used.

$P\left(E_1\right)=\frac{1}{10}$, 

$P\left(E_2\right)=\frac{9}{10}$

Let E be the event that it collapses

$P\left(E\mid E_1\right)=\frac{95}{100}$

$P\left(E\mid E_2\right)=\frac{45}{100}$, 

Required probability = $P\left(E_1\mid E\right)$

= Probability that the bridge collapses and the chance that it was due to wrong design.

$P(E1|E)=\frac{P\left(E_1\right)\cdot P\left(E\mid E_1\right)}{P\left(E_1\right)\cdot P\left(E\mid E_2\right)+P\left(E_2\right)\cdot P\left(E\mid E_2\right)}$

$=\frac{\left(\frac{1}{10}\times \frac{95}{100}\right)}{\left(\frac{1}{10}\times \frac{95}{100}\right)+\left(\frac{9}{10}\times \frac{45}{100}\right)}$

$=\frac{\left(\frac{95}{1000}\right)}{\left(\frac{95}{1000}\right)+\left(\frac{9\times 45}{1000}\right)}$

$=\frac{95}{95+405}$

$=\frac{95}{500}=\frac{19}{100}=0.19$

Hence the required probability is 0.19

Example 5: A man is known to speak the 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.

Solution: 

In a roll of a die, let

E₁ = event of rolling a six,

E₂ = event of not rolling a six, and

E = event that the man reports the result as a six.

Then, $P\left(E_1\right)=\frac{1}{6}$, and 

$P\left(E_2\right)=\left(1-\frac{1}{6}\right)=\frac{5}{6}$

P(E|E1) is the probability that the man reports a six occurs, when six has actually occurred.

= probability that the man speaks the truth

$=\frac{3}{4}$

P(E|E2) = probability that the man report that six occurs, when six has not actually occurred

= probability that the man does not speak the truth

$=\left(1-\frac{3}{4}\right)=\frac{1}{4}$

Probability of getting a six given that the man reports it to be six

$=P\left(E_1\mid E\right)$

$=\frac{P\left(E_1\mid E\right)\cdot P\left(E_1\right)}{P\left(E\mid E_1\right)\cdot P\left(E_1\right)+P\left(E\mid E_2\right)\cdot P\left(E_2\right)}$

[By Bayes’ theorem]

$=\frac{\left(\frac{3}{4}\times \frac{1}{6}\right)}{\left(\frac{3}{4}\times \frac{1}{6}\right)+\left(\frac{1}{4}\times \frac{5}{6}\right)}$

$=\left(\frac{1}{8}\times 3\right)=\frac{3}{8}$

Hence, the required probability is $\frac{3}{8}.$

Example 6: A card is lost from a standard deck of 52 cards. From the remaining 51 cards remaining in the deck, two cards are drawn, and both are found to be spades. What is the probability that the lost card was a spade?

Solution: 

Let E₁, E₂, E₃ and E₄ represent the events of losing a card of spades, clubs, hearts and diamonds respectively.

Then, $P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=P\left(E_4\right)=\frac{13}{52}=\frac{1}{4}$

Let E be the event of drawing 2 spades from the remaining 51 cards. Then,

P(E|E₁) = Probability of drawing 2 spades, given that a card from the clubs suit is missing.

$=\frac{{}^{12}{\mathrm{C}}_2}{{}^{51}{\mathrm{C}}_2}=\frac{(12\times 11)}{2!}\times \frac{2!}{(51\times 50)}=\frac{22}{425}$

P(E|E₂) = Probability of drawing 2 spades, given that a card of clubs is missing.

$=\frac{{}^{13}{\mathrm{C}}_2}{{}^{51}{\mathrm{C}}_2}=\frac{(13\times 12)}{2!}\times \frac{2!}{(51\times 50)}=\frac{26}{425}$

P(E|E₃) = Probability of drawing 2 spades from the remaining cards, given that a card of hearts is missing.

$=\frac{{}^{13}{\mathrm{C}}_2}{{}^{51}{\mathrm{C}}_2}=\frac{26}{425}$

P(E|E₄) = Probability of drawing 2 spades, given that a card of diamonds is missing.

$=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{26}{425}$

P(E₁|E) = Probability of the lost card being a spade, given that 2 spades are drawn from the remaining 51 cards.

$=\frac{P\left(E_1\right)\cdot P\left(E\mid E_1\right)}{P\left(E_1\right)\cdot P\left(E\mid E_1\right)+P\left(E_2\right)\cdot P\left(E\mid E_2\right)+P\left(E_3\right)\cdot P\left(E\mid E_3\right)+P\left(E_4\right)\cdot P\left(E\mid E_4\right)}$

$=\frac{\left(\frac{1}{4}\times \frac{22}{425}\right)}{\left(\frac{1}{4}\times \frac{22}{425}\right)+\left(\frac{1}{4}\times \frac{26}{425}\right)+\left(\frac{1}{4}\times \frac{26}{425}\right)+\left(\frac{1}{4}\times \frac{26}{425}\right)}$

$=\frac{22}{100}=0.22$

Hence, the required probability is 0.22

4.0Bayes Theorem Questions for Practice

1.  A doctor is scheduled to visit a patient, and based on past experience, the probabilities that he will use a train, bus, scooter, or other means of transport are respectively $\frac{3}{10},\frac{1}{5},\frac{1}{10}$ and $\frac{2}{5}$. The probabilities of him being late if he travels by train, bus, or scooter are also $\frac{1}{4},\frac{1}{3}$ and $\frac{1}{12}$. However, if he uses other means of transport, he will not be late. Given that he is late upon arrival, what is the probability that he traveled by train?
2. If a machine is set up correctly, it produces 90% acceptable items. However, if it is set up incorrectly, only 40% of the items are acceptable. Based on past experience, 80% of the machine set-ups are correct. Given that the machine produces 2 acceptable items, what is the probability that the machine was set up correctly?
3. A company manufactures scooters at two plants, A and B. Plant A produces 80% of the total scooters, while Plant B produces 20%. Of the scooters produced, 85% from Plant A and 65% from Plant B are of standard quality. If a scooter is randomly selected and found to be of standard quality, what is the probability that it was produced at Plant A?
4. Bag A contains 1 white ball and 6 red balls, while another bag contains 4 white balls and 3 red balls. One of the bags is randomly selected, and a ball is drawn from it, which turns out to be white. What is the probability that the white ball was drawn from Bag A?
5. Three urns—A, B, and C—contain the following balls: Urn A has 6 red and 4 white balls, Urn B has 2 red and 6 white balls, and Urn C has 1 red and 5 white balls. An urn is randomly chosen, and a ball is drawn, which turns out to be red. What is the probability that this red ball was drawn from Urn A?
6. Urn A contains 6 white and 4 black balls, Urn B contains 4 white and 6 black balls, and Urn C contains 2 white and 8 black balls. Each urn is selected at random with probabilities of 0.2, 0.6, and 0.2, respectively. Two balls are drawn without replacement from the chosen urn, and both are found to be white. What is the probability that these balls were drawn from Urn C?
7. There are four boxes, A, B, C, and D, containing marbles. A contains 1 red, 6 white and 3 black marbles; B contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?

5.0Sample Questions on Bayes’ Theorem Questions

1. How do you solve Bayes' Theorem probability questions?  

Ans: To solve Bayes' Theorem probability questions, follow these steps:

• Identify the prior probability (P(A)) and likelihood (P(B|A)).
• Determine the marginal likelihood (P(B)) using the law of total probability.
• Apply the formula $P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$ to calculate the posterior probability.

2. What is the distinction between prior probability and posterior probability?  

Ans: To calculate posterior probability, you need the prior probability (P(A)), the likelihood (P(B|A)), and the marginal likelihood (P(B)). Use Bayes' Theorem:

$P(A\mid B)=\frac{P(B\mid A)\cdot P(A)}{P(B)}$

This gives the updated probability of A after considering evidence B.