## Bayes Theorem and Its Applications

**Theorem of Total Probability**

**Theorem:** Let E_{1}, E_{2},..E_{n} be mutually exclusive and exhaustive events associated with a random experiment and let E be an event that occurs with some E_{i}. Then, prove that

⇒ P(E) = \(\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right) .\)

**Example: There are three urns containing 3 white and 2 black balls; 2 white and 3 black balls; 1 black and 4 white balls respectively. There is an equal probability of each urn being chosen. One ball is drawn from an urn chosen at random. What is the probability that a white ball is drawn?**

**Solution:**

**Read and Learn More WBCHSE Solutions For Class 12 Maths**

Let E_{1}, E_{2,} and E_{3} be the events of choosing the first, second, and third urn respectively.

Then, P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{3}\)

Let E be the event that a white ball is drawn. Then,

P(E/E_{1}) = \(\frac{3}{5}\), P(E/E_{2}) = \(\frac{2}{5}\) and P(E/E_{3}) = \(\frac{4}{5}\)

By the theorem of total probability, we have

P(E) = P(E/E_{1}) • P(E_{1}) + P(E/E_{2}) • P(E_{2}) + P(E/E_{3}) • P(E_{3})

= \(\left(\frac{3}{5} \times \frac{1}{3}+\frac{2}{5} \times \frac{1}{3}+\frac{4}{5} \times \frac{1}{3}\right)=\left(\frac{1}{5}+\frac{2}{15}+\frac{4}{15}\right)=\frac{9}{15}=\frac{3}{5} .\)

**Bayes’ Theorem: **Let E_{1}, E_{2}…, E_{n} be mutually exclusive and exhaustive events, associated with a random experiment, and let E be any event that occurs with some Ei. Then,

⇒ \(P\left(E_i / E\right)=\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)} ; i=1,2,3, \ldots, n\)

**Proof:** By the theorem of total probability, we have

P(E) = \(\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)\)….(1)

∴ \(P\left(E_i / E\right)= \left.\frac{P\left(E \cap E_i\right)}{P(E)} \quad \text { [by multiplication theorem }\right]\)

= \(\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{P(E)} \quad\left[because P\left(E / E_i\right)=\frac{P\left(E \cap E_i\right)}{P\left(E_i\right)}\right]\)

= \(\left.\frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)} \text { [using }(\mathrm{i})\right]\)

Hence, \(P\left(E_i / E\right)= \frac{P\left(E / E_i\right) \cdot P\left(E_i\right)}{\sum_{i=1}^n P\left(E / E_i\right) \cdot P\left(E_i\right)}\).

## Bayes Theorem Solved Examples

**Example 1. A factory has three machines, X, Y, and Z, producing 1000,2000 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and it is found to be defective. What is the probability that this defective bolt has been produced by the machine X?**

**Solution:**

Total number of bolts produced in a day = (1000 + 2000 + 3000) = 6000.

Let E_{1}, E_{2,} and E_{3} be the events of drawing a bolt produced by machines X, Y, and Z respectively.

Then, \(P\left(E_1\right)=\frac{1000}{6000}=\frac{1}{6} ; P\left(E_2\right)=\frac{2000}{6000}=\frac{1}{3} \text { and } P\left(E_3\right)=\frac{3000}{6000}=\frac{1}{2}\)

Let E be the event of drawing a defective bolt.

Then, P(E/E_{1}) = probability of drawing a defective bolt, given that it is produced by the machine X = \(\frac{1}{100}\)

P(E/E_{2}) = probability of drawing a defective bolt, given that it is produced by the machine Y = \(\frac{1.5}{100}\)= \(\frac{15}{1000}\) = \(\frac{3}{200}\)

P(E/E_{3}) = probability of drawing a defective bolt, given that it is produced by the machine Z = \(\frac{2}{100}\) = \(\frac{1}{50}\)

Required probability = P(E_{1}/E) = probability that the bolt drawn is produced by X, given that it is defective

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

= \(\frac{\left(\frac{1}{6} \times \frac{1}{100}\right)}{\left(\frac{1}{6} \times \frac{1}{100}\right)+\left(\frac{1}{3} \times \frac{3}{200}\right)+\left(\frac{1}{2} \times \frac{1}{50}\right)}\)

= \(\left(\frac{1}{600} \times \frac{600}{10}\right)=\frac{1}{10}=0.1\)

Hence, the required probability is 0.1.

**Example 2. In a bolt factory, three machines, A, B, and C, manufacture 25%, 35%, and 40% of the total production respectively. Of their respective outputs, 5%, 4%, and 2% are defective. A bolt is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by machine C.**

**Solution:**

Let E_{1}, E_{2,} and E_{3} be the events of drawing a bolt produced by machines A, B, and C respectively.

Then \(P\left(E_1\right)=\frac{25}{100}=\frac{1}{4}, P\left(E_2\right)=\frac{35}{100}=\frac{7}{20} \text {, and } P\left(E_3\right)=\frac{40}{100}=\frac{2}{5} \text {. }\)

Let E be the event of drawing a defective bolt. Then,

P(E/E_{1}) = probability of drawing a defective bolt, given that it is produced by the machine A = \(\frac{5}{100}\)=\(\frac{1}{20}\)

P(E/E_{2}) = probability of drawing a defective bolt, given that it is produced by the machine B = \(\frac{4}{100}\)=\(\frac{1}{25}\)

P(E/E_{3}) = probability of drawing a defective bolt, given that it is produced by the machine C = \(\frac{2}{100}\)=\(\frac{1}{50}\)

The probability that the bolt drawn is manufactured by C, given that it is defective

= \(P\left(E_3 / E\right)\)

= \(\frac{P\left(E / E_3\right) \cdot P\left(E_3\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{1}{50} \times \frac{2}{5}\right)}{\left(\frac{1}{20} \times \frac{1}{4}\right)+\left(\frac{1}{25} \times \frac{7}{20}\right)+\left(\frac{1}{50} \times \frac{2}{5}\right)}=\left(\frac{1}{125} \times \frac{2000}{69}\right)=\frac{16}{69}\) .

Hence, the required probability is \(\frac{16}{69}\)

**Example 3. A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant, 40%. Also, 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality. Find the probability that it comes from the second plant.**

**Solution:**

Let E_{1} and E_{2} be the events of choosing a bicycle from the first plant and the second plant respectively.

Then, \(P\left(E_1\right)=\frac{60}{100}=\frac{3}{5} \text {, and } P\left(E_2\right)=\frac{40}{100}=\frac{2}{5} \text {. }\)

Let E be the event of choosing a bicycle of standard quality. Then,

P(E/E_{1}) = probability of choosing a bicycle of standard quality, given that it is produced by the first plant = \(\frac{80}{100}\) = \(\frac{4}{5}\)

P(E/E_{2}) = probability of choosing a bicycle of standard quality, given that it is produced by the second plant = \(\frac{90}{100}\) = \(\frac{9}{10}\)

The required probability

P(E_{2}/E) = probability of choosing a bicycle from the second plant, given that it is of standard quality

= \(\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{2}{5} \times \frac{9}{10}\right)}{\left(\frac{3}{5} \times \frac{4}{5}\right)+\left(\frac{2}{5} \times \frac{9}{10}\right)}=\frac{3}{7}\).

**Example 4. An insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probability of an accident involving a scooter, a car, and a truck is 1/ 100, 3/100, and 3/20 respectively. One of the insured persons met with an accident. What is the probability that he is a scooter driver?**

**Solution:**

Total number of persons insured = (2000 + 4000 + 6000) = 12000.

Let E_{1}, E_{2,} and E_{3} be the events of choosing a scooter driver, a car driver, and a truck driver respectively.

Then, \(P\left(E_1\right)=\frac{2000}{12000}=\frac{1}{6}, P\left(E_2\right)=\frac{4000}{12000}=\frac{1}{3} \text {, and } P\left(E_3\right)=\frac{6000}{12000}=\frac{1}{2} \text {. }\)

Let E be the event of an insured person meeting with an accident. Then,

P(E/E_{1}) = probability that an insured person meets with an accident, given that he is a scooter driver = \(\frac{1}{100}|\)

Similarly, P(E/E_{2}) = \(\frac{3}{100}|\) and P(E/E_{3}) = \(\frac{3}{20}|\)

Required probability = P(E_{1}/E) [by Bayes’ theorem]

= probability of choosing a scooter driver, given that he meets with an accident

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

= \(\frac{\left(\frac{1}{100} \times \frac{1}{6}\right)}{\left(\frac{1}{100} \times \frac{1}{6}\right)+\left(\frac{3}{100} \times \frac{1}{3}\right)+\left(\frac{3}{20} \times \frac{1}{2}\right)}=\frac{1}{52}\).

Hence, the required probability is \(\frac{1}{52}\)

**Example 5. A doctor to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter, or by car are respectively \(\frac{3}{100}\), \(\frac{1}{5}\), \(\frac{1}{10}\) and \(\frac{2}{5}\). The probabilities that he will be late are \(\frac{1}{4}\), \(\frac{1}{3}\), and \(\frac{1}{12}\), if he comes by train, bus and scooter respectively; but if he comes by car, he will not be late. When he arrives, he is late. What is the probability that he has come by train?**

**Solution:**

Let E_{1}, E_{2}, E_{3}, and E_{4} be the events that the doctor comes by train, bus, scooter, and car respectively.

Then, \(P\left(E_1\right)=\frac{3}{10}, P\left(E_2\right)=\frac{1}{5}, P\left(E_3\right)=\frac{1}{10} \text { and } P\left(E_4\right)=\frac{2}{5} \text {. }\)

Let E be the event that the doctor is late. Then,

P(E/E_{1}) = probability that the doctor is late, given that he comes by train = \(\frac{1}{4}\)

P(E/E_{2}) = probability that the doctor is late, given that he comes by bus =\(\frac{1}{3}\)

P(E/E_{3}) = probability that the doctor is late, given that he comes by scooter = \(\frac{1}{12}\)

P(E/E_{4}) = probability that the doctor is late, given that he comes by car = 0.

The probability that he comes by train, given that he is late = P(E_{1}/E)

= \(\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)+P\left(E_4\right) \cdot P\left(E / E_4\right)}\) [by Bayes’ theorem]

\(\left.=\frac{\left(\frac{3}{10} \times \frac{1}{4}\right)}{\left(\frac{3}{10} \times \frac{1}{4}\right)+\left(\frac{1}{5} \times \frac{1}{3}\right)+\left(\frac{1}{10} \times \frac{1}{12}\right)+\left(\frac{2}{5} \times 0\right)} \times \frac{120}{18}\right)=\frac{1}{2}\).

Hence the required probability is \(\frac{1}{2}\)

**Example 6. 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 throw of a die, let

E_{1} = event of getting a six,

E_{2} = event of not getting a six, and

E = event that the man. reports that it is a six. Then,

Then, P(E_{1}) = \(\frac{1}{6}\) , and P(E_{2}) = (1-\(\frac{1}{6}\) = \(\frac{1}{5}[/latex

P(E/E_{1}) = probability that tire man reports that six occurs when six has actually occurred

= probability that the man speaks the truth = [latex]\frac{3}{4}\)

P(E/E_{2}) = probability that the man reports that six occurs when six has not actually occurred

= probability that the man does not speak the truth = (1-\(\frac{3}{4}\)) = \(\frac{1}{4}\)

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

= \(P\left(E_1 / E\right)\)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\) [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 7. In an examination, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices. The probability that he makes a guess is (1/3) and the probability that he copies the answer is (1/6). The probability that his answer is correct, given that he copied it, is (1/8). The probability that his answer is correct, given that he guessed it, is (1/4). Find the probability that he him the answer to the question, given that he correctly answered it.**

**Solution:**

Let E_{1} = event that the examinee guesses the answer,

E_{2} = event that he copies the answer,

E_{3} = event that he knows the answer, and

E = event that he answers correctly.

Then, \(P\left(E_1\right)=\frac{1}{3}, P\left(E_2\right)=\frac{1}{6}, \text { and } P\left(E_3\right)=1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}\)

[E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive].

∴ P(E/E_{1}) = probability that he answers correctly, given that he guesses \(\frac{1}{4}\)

P(E/E_{2}) = probability that he answers correctly, given that he copies \(\frac{1}{8}\)

P(E/E_{3}) = probability that he answers correctly, given that he knew the answer = 1.

Required probability = \(P\left(E_3 / E\right)\)

= \(\frac{P\left(E / E_3\right) \cdot P\left(E_3\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)}\) [by Bayes’ theorem]

= \(\frac{\left(1 \times \frac{1}{2}\right)}{\left(\frac{1}{4} \times \frac{1}{3}\right)+\left(\frac{1}{8} \times \frac{1}{6}\right)+\left(1 \times \frac{1}{2}\right)}=\frac{24}{29}\).

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

**Example 8. By examining the chest X-ray, the probability that a person is diagnosed with TB when he is actually suffering from it is 0.99. The probability that the doctor incorrectly diagnoses a person to be having TB, on the basis of X-ray reports, is 0.001. In a certain city, 1 in 1000 persons suffers from TB. A person is selected at random and is diagnosed to have TB. What is the chance that he actually has TB?**

**Solution:**

Let E = event that the doctor diagnoses TB,

E_{1} = event that the person selected is suffering from TB, and

E_{2} = event that the person selected is not suffering from TB.

P(E/E_{1}) = probability that TB is diagnosed when the person actually has TB = \(\frac{99}{100}\)

P(E/E_{2}) = probability that TB is diagnosed, when the person has no TB = \(\frac{1}{1000}\)

Using Bayes’ theorem, we have

P(E_{1}/E) = probability of a person actually having TB, if it is known that he is diagnosed to have TB

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

= \(\frac{\left(\frac{99}{100} \times \frac{1}{1000}\right)}{\left(\frac{99}{100} \times \frac{1}{1000}\right)+\left(\frac{1}{1000} \times \frac{999}{1000}\right)}=\frac{110}{221}\).

Hence, the required probability is \(\frac{110}{221}\).

**Example 9. Bag A contains 2 white and 3 red balls, and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag B.**

**Solution:**

Let E_{1} = event of choosing bag A,

E_{2} = event of choosing bag B, and E = event of drawing a red ball.

Then, P(E_{1}) = \(\frac{1}{2}\) and P(E_{2}) = \(\frac{1}{2}\)

Also, P(E/E_{1}) = event of drawing a red ball from bag A = \(\frac{3}{5}\) and

P(E/E_{2}) = event of drawing a red ball from bag B = \(\frac{5}{9}\)

The probability of drawing a ball from B, it is given that it is red = P(E_{2}/E)

= \(\frac{P\left(E / E_2\right) \cdot P\left(E_2\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{5}{9} \times \frac{1}{2}\right)}{\left(\frac{3}{5} \times \frac{1}{2}\right)+\left(\frac{5}{9} \times \frac{1}{2}\right)}=\frac{25}{52}\).

Hence, the required probability is \(\frac{25}{52}\).

**Example 10. There are 5 hags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.**

**Solution:**

Let E_{1} = event of selecting a bag from the first group,

E_{2} = event of selecting a bag from the second group, and

E = event of drawing a white ball.

Then, P(E_{1}) = \(\frac{5}{11}\) and P(E_{2}) = \(\frac{6}{11}\)

P(E/E_{1}) = probability of getting a white ball, given that it is from a bag of the first group = \(\frac{5}{8}\)

P(E/E_{2}) = probability of getting a white ball, given that it is from a bag of the second group = \(\frac{2}{6}\)= \(\frac{1}{3}\)

The probability of getting the ball from a bag of the first group, given that it is white = P(E_{1}/E)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\) [by Bayes’ theorem]

= \(\frac{\left(\frac{5}{8} \times \frac{5}{11}\right)}{\left(\frac{5}{8} \times \frac{5}{11}\right)+\left(\frac{1}{3} \times \frac{6}{11}\right)}=\frac{75}{123}\).

**Example 11. Urn A contains 1 white, 2 black, and 3 red balls; urn B contains 2 white, 1 black, and 1 red ball; and urn C contains 4 white, 5 black, and 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red. What is the probability that they come from urn A?**

**Solution:**

Let E_{1}, E_{2}, and E_{3} be the events that the balls are drawn from urn A, urn B, and urn C respectively, and let E be the event that the balls drawn are one white and one red.

Then, P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{5}\)

P(E/E_{1}) = probability that the balls drawn are one white and one red, given that the balls are from urn A

= \(\frac{{ }^1 C_1 \times{ }^3 C_1}{{ }^6 C_2}=\frac{3}{15}=\frac{1}{5}\)

P(E/E_{2}) = probability that the balls drawn are one white and one red, given that the balls are from urn B

= \(\frac{{ }^2 C_1 \times{ }^1 C_1}{{ }^4 C_2}=\frac{2}{6}=\frac{1}{3} .\)

P(E/E_{3}) = probability that the balls drawn are one white and one red, given that the balls are from urn C

= \(\frac{{ }^4 C_1 \times{ }^3 C_1}{{ }^{12} C_2}=\frac{12}{66}=\frac{2}{11}\)

The probability that the balls drawn are from urn A, it is given that the balls drawn are one white and one red = P(E_{1}/E)

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

[by Bayes’ theorem]

= \(\frac{\left(\frac{1}{5} \times \frac{1}{3}\right)}{\left(\frac{1}{5} \times \frac{1}{3}\right)+\left(\frac{1}{3} \times \frac{1}{3}\right)+\left(\frac{2}{11} \times \frac{1}{3}\right)}\)

= \(\left(\frac{1}{15} \times \frac{495}{118}\right)=\frac{33}{118}\).

Hence, the required probability is \(\frac{33}{118}\)

**Example 12. A card from a pack of 52 cards is lost. From the remaining cards of the -pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.**

**Solution:**

Let E_{1}, E_{2}, E_{3,} and E_{4} be the events of losing a card of spades, clubs, hearts, and diamonds respectively.

Then, P(E_{1}) = P(E_{2}) = P(E_{3}) = P(E_{4}) = \(\frac{13}{52}\) = \(\frac{1}{4}\)

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

P(E/E_{1}) = probability of drawing 2 spades, given that a card of spades is missing

= \(\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{(12 \times 11)}{2 !} \times \frac{2 !}{(51 \times 50)}=\frac{22}{425}\).

P(E/E_{2}) = probability of drawing 2 spades, given that a card of clubs is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{(13 \times 12)}{2 !} \times \frac{2 !}{(51 \times 50)}=\frac{26}{425}\).

P(E/E_{3}) = probability of drawing 2 spades, given that a card of hearts is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\)

P(E/E_{4}) = probability of drawing 2 spades, given that a card of diamonds is missing

= \(\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\)

∴ P(E_{1}/E) = probability of the lost card being a spade, given that 2 spades are drawn from the remaining 51 cards

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

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

= \(\frac{22}{100}\)=0.22.

Hence, the required probability is 0.22.

## Bayes Theorem Exercise

**Question 1. In a bulb factory, three machines, A, B, and C, manufacture 60%, 25%, and 15% of the total production respectively. Of their respective outputs, 1%, 2%, and 1% are defective. A bulb is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by machine C.**

**Solution:** 2/25

**Question 2. A company manufactures scooters at two plants, A and B. Plant A produces 80% and plant B produces 20% of the total product. 85% of the scooters produced at plant A and 65% of the scooters produced at plant B are of standard quality. A scooter produced by the company is selected at random and it is found to be of standard quality. What is the probability that it was manufactured at plant A.**

**Solution: **68/81

**Hint**: Let E_{1} = event that the selected scooter is produced at plant A, and

E_{2} = event that the selected scooter is produced at plant B.

Then, P(E_{1}) = \(\frac{80}{100}\) = \(\frac{4}{5}\)and P(E_{2}) = \(\frac{20}{100}\) = \(\frac{1}{5}\)

Let E be the event of choosing a scooter that is of standard quality.

The probability that the selected scooter’s standard quality

Then, \(P\left(E / E_1\right)=\frac{85}{100}=\frac{17}{20} \text {, and } P\left(E / E_2\right)=\frac{65}{100}=\frac{13}{20}\)

The probability that the selected scooter was produced at plant A, given that is is of standard quality = \(P\left(E_1 / E\right)\)

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

**Question 3. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students are girls. If a student is selected at random and is taller than 1.75 meters, what is the probability that the selected student is a girl?**

**Solution:** 3/11

**Hint:** Let E_{1} and E_{2} be the events of selecting a boy and a girl respectively.

Then, P(E_{1}) = \(\frac{40}{100}\) = \(\frac{2}{5}\), and P(E_{2}) = \(\frac{60}{100}\) = \(\frac{3}{5}\)

Let E = event that the student selected is taller than 1.75 m.

Then, \(P\left(E / E_1\right)=\frac{4}{100}=\frac{1}{25} \text { and } P\left(E / E_2\right)=\frac{1}{100}\)

The probability that the selected student is a girl, given that she is taller than 1.75 m. = \(P\left(E_2 / E\right)\)

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

**Question 4. In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In this class, 60% of the students are boys. If a student is selected at random and found to have an IQ of more than 150, find the probability that the student is a boy.**

**Solution:** 3/7

**Hint:** Let E_{1} and E_{2} be the events of selecting a boy and a girl respectively. Then,

⇒ \(P\left(E_1\right)=\frac{60}{100}=\frac{3}{5} \text {, and } P\left(E_2\right)=\frac{40}{100}=\frac{2}{5} \text {. }\)

Let E be the event of selecting a student having an IQ of more than 150.

Then, \(P\left(E / E_1\right)=\frac{5}{100}=\frac{1}{20} \text {, and } P\left(E / E_2\right)=\frac{10}{100}=\frac{1}{10} \text {. }\)

Required probability = P(E_{1}/E)

= \(=\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)} .\)

**Question 5. Suppose 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.**

**Solution:** 2/3

**Hint:** Let there be 1000 males and 1000 females.

Let E_{1} and E_{2} be the events of choosing a male and a female respectively.

Then, \(P\left(E_1\right)=\frac{1000}{2000}=\frac{1}{2} \text {, and } P\left(E_2\right)=\frac{1000}{2000}=\frac{1}{2} \text {. }\)

Let E be the event of choosing a grey-haired person. Then,

⇒ \(P\left(E / E_1\right)=\frac{50}{1000}=\frac{1}{20} \text {, and } P\left(E / E_2\right)=\frac{25}{1000}=\frac{1}{40} \text {. }\)

The probability of selecting a male person, given that the person selected is a grey-haired = P(E_{1}/E)

= \(\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)} .\)

**Question 6. Two groups are competing for the positions on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and when the second group wins, the corresponding probability is 0.3. Find the probability that the new product introduced was by the second group.**

**Solution:** 2/9

**Hint:** Let E_{1} = event that the first group wins,

E_{2} = event that the second group wins, and

E = event that a new product is introduced.

Then, P(E_{1}) = 0.6, P(E_{2}) = 0.4,

P(E/E_{1}) = 0.7, P(E/E_{2}) = 0.3.

Required probability = P(E_{2}/E)

=\(\frac{P\left(E / E_2\right) \cdot P\left(E_2\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)} .\)

**Question 7. A bag A contains 1 white and 6 red balls. Another bag contains 4 white and 3red balls. One of the bags is selected at random and a ball is drawn from it, which is found to be white. Find the probability that the ball drawn is from bag A.**

**Solution:** 1/5

**Question 8. There are two bags 1 and 2. Bag 1 contains 3 white and 4 black balls, and bag 2 contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it was drawn from bag 1.**

**Solution:** 33/88

**Question 9. A box contains 2 gold and 3 silver coins. Another box contains 3 gold and 3 silver coins. A box is chosen at random, and a coin is drawn from it. If the selected coin is a gold coin, find the probability that it was drawn from the second box.**

**Solution**: 5/9

**Question 10. Three urns A, B, and C contain 6 red and 4 white; 2 red and 6 white; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from the urn A.**

**Solution:** 36/61

**Hint:** Let E_{1}, E_{2,} and E_{3} be the events of choosing the urns A, B, and C respectively, and let £ be the event of drawing a red ball. Then,

P(E_{1}) = P(E_{2}) = P(E_{3}) = 1/3

⇒ \(P\left(\frac{E}{E_1}\right)=\frac{6}{10}, P\left(\frac{E}{E_2}\right)= \frac{2}{8} \text {, and } P\left(\frac{E}{E_3}\right)=\frac{1}{6}\)

Required probability = \(P\left(\frac{E_1}{E}\right)\)

= \(\frac{P\left(E_1\right) \times P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) \times P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \times P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \times P\left(\frac{E}{E_3}\right)}\)

= \(\frac{\left(\frac{1}{3} \times \frac{6}{10}\right)}{\left(\frac{1}{3} \times \frac{6}{10}\right)+\left(\frac{1}{3} \times \frac{2}{8}\right)+\left(\frac{1}{3} \times \frac{1}{6}\right)}=\frac{36}{61}\)

**Question 11. Three urns contain 2 white and 3 black balls; 3 white and 2 black balls, and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn.**

**Solution:** 2/9

**Question 12. There are three boxes, the first one containing 1 white, 2 red, and 3 black balls; the second one containing 2 white, 3 red, and 1 black ball and the third one containing 3 white, 1 red, and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box?**

**Solution:** 6/11

**Question 13. Urn A contains 7 white and 3 black balls; urn B contains 4 white and 6 black balls; urn C contains 2 white and 8 black balls. One of these urns is chosen at random with probabilities 0.2, 0.6 and 0.2 respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls drawn are from the urn C.**

**Solution:** 1/40

**Hint:** P(A) = 0.2, P(B) = 0.6 and P(Q) = 0.2.

Let E be the event that 2 white balls are drawn. Then,

P(E/A)= \(\frac{{ }^7 C_2}{{ }^{10} C_2} ; P(E / B)=\frac{{ }^4 C_2}{{ }^{10} C_2} ; P(E / C)=\frac{{ }^2 C_2}{{ }^{10} C_2}\)

∴ required probability = P(C/E)

= \(\frac{P(E / C) \cdot P(C)}{P(E / A) \cdot P(A)+P(E / B) \cdot P(B)+P(E / C) \cdot P(C)} \text {. }\)

**Question 14. There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.**

**Solution:** 45/61

**Hint:** Let E_{1} = event of selecting a bag from the first group, and

E_{2} = event of selecting a bag from the second group.

Then, P(E_{1}) = 3/5 and P(E_{2}) = 2/5

Let E = event that the ball drawn is white. Then,

P(E/E_{1}) = 5/8, P(E/E_{2}) = 2/6 = 1/3

⇒ \(P\left(E_1 / E\right)=\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)}\)

**Question 15. 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?**

**Solution:** 1/15

**Hint:** Let E_{1}, E_{2}, E_{3}, E_{4} be the events of selecting boxes A, B, C, and D respectively. Then,

P(E_{1}) = P(E_{2}) = P(E_{3}) = P(E_{4}) = 1/4

Let E = event that the marble drawn is red. Then,

P(E/E_{1}) = 1/10, P(E/E_{2}) = 6/10 = 3/5, P(E/E_{3}) = 8/10, 4/5, P(E/E_{4}) =0

∴ \(P\left(E_1 / E\right)=\frac{P\left(E / E_1\right) \cdot P\left(E_1\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) \cdot P\left(E_2\right)+P\left(E / E_3\right) \cdot P\left(E_3\right)+P\left(E / E_4\right) \cdot P\left(E_4\right)} .\)

**Question 16. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and Plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X.**

**Solution:** 56/83

**Hint:** Let E_{1} and E_{2} be the events that the car is manufactured by plants X and Y respectively.

Let E be the event that the car is of standard quality. Then

⇒ \(P\left(E_1\right)=\frac{70}{100}=\frac{7}{10}, P\left(E_2\right)=\frac{30}{100}=\frac{3}{10}\) ;

⇒ \(P\left(E / E_1\right)=\frac{80}{100}=\frac{4}{5}, P\left(E / E_2\right)=\frac{90}{100}=\frac{9}{10}\)

∴ \(P\left(E_1 / E\right)=\frac{P\left(E_1\right) \times P\left(E / E_1\right)}{P\left(E_1\right) \times P\left(E / E_1\right)+P\left(E_2\right) \times P\left(E / E_2\right)}\)

= \(\frac{\left(\frac{7}{10} \times \frac{4}{5}\right)}{\left(\frac{7}{10} \times \frac{4}{5}\right)+\left(\frac{3}{10} \times \frac{9}{10}\right)}=\frac{56}{83}\)

**Question 17. An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accident vehicle was a motorcycle.**

**Solution:** 3/4

**Hint:** Let E_{1} and E_{2} be the events that an insured vehicle is a scooter and a motorcycle respectively.

Let E be the event that the insured vehicle meets an accident.

⇒ \(P\left(E_1\right)=\frac{2000}{(2000+3000)}=\frac{2}{5}, P\left(E_2\right)=\frac{3000}{5000}=\frac{3}{5}\)

⇒ \(P\left(E / E_1\right)=0.01 \text { and } P\left(E / E_2\right)=0.02\)

∴ \(P\left(E_2 / E\right)=\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\)

= \(\frac{\left(\frac{3}{5} \times 0.02\right)}{\left(\frac{2}{5} \times 0.01\right)+\left(\frac{3}{5} \times 0.02\right)}=\frac{3}{4}\)

**Question 18. In a bulb factory, machines A, B, and C manufacture 60%, 30%, and 10% of bulbs respectively. Out of these bulbs 1%, 2%, and 3% of the bulbs produced respectively by A, B, and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by machine A.**

**Solution:** 2/5