This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Bayes’ Theorem”.

1. Three companies A, B and C supply 25%, 35% and 40% of the notebooks to a school. Past experience shows that 5%, 4% and 2% of the notebooks produced by these companies are defective. If a notebook was found to be defective, what is the probability that the notebook was supplied by A?

a) ^{44}⁄_{69}

b) ^{25}⁄_{69}

c) ^{13}⁄_{24}

d) ^{11}⁄_{24}

View Answer

Explanation: Let A, B and C be the events that notebooks are provided by A, B and C respectively.

Let D be the event that notebooks are defective

Then,

P(A) = 0.25, P(B) = 0.35, P(C) = 0.4

P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02

P(A│D) = (P(D│A)*P(A))/(P(D│A) * P(A) + P(D│B) * P(B) + P(D│C) * P(C) )

=(0.05*0.25)/((0.05*0.25)+(0.04*0.35)+(0.02*0.4)) = 2000/(80*69)

= ^{25}⁄_{69}.

2. A box of cartridges contains 30 cartridges, of which 6 are defective. If 3 of the cartridges are removed from the box in succession without replacement, what is the probability that all the 3 cartridges are defective?

View Answer

Explanation: Let A be the event that the first cartridge is defective. Let B be the event that the second cartridge is defective. Let C be the event that the third cartridge is defective. Then probability that all 3 cartridges are defective is P(A ∩ B ∩ C)

Hence,

P(A ∩ B ∩ C) = P(A) * P(B|A) * P(C | A ∩ B)

= (

^{6}⁄

_{30}) * (

^{5}⁄

_{29}) * (

^{4}⁄

_{28})

= ^{(6 * 5 * 4)}⁄_{(30 * 29 * 28)}.

3. Two boxes containing candies are placed on a table. The boxes are labelled B_{1} and B_{2}. Box B^{1} contains 7 cinnamon candies and 4 ginger candies. Box B_{2} contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B_{1} is ^{1}⁄_{3} and the probability of selecting box B_{2} is ^{2}⁄_{3}. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. What is the probability that Suresh will win the TV (that is, she will select a cinnamon candy)?

a) ^{7}⁄_{33}

b) ^{6}⁄_{33}

c) ^{13}⁄_{33}

d) ^{20}⁄_{33}

View Answer

Explanation: Let A be the event of drawing a cinnamon candy.

Let B1 be the event of selecting box B1.

Let B2 be the event of selecting box B2.

Then, P(B1) =^{1⁄3} and P(B2) = ^{2⁄3}

P(A) = P(A ∩ B1) + P(A ∩ B2)

= P(A|B1) * P(B1) + P(A|B2)*P(B2)

= (^{7}⁄_{11}) * (^{1}⁄_{3}) + (^{3}⁄_{11}) * (^{2}⁄_{3})

= ^{13}⁄_{33}.

4. Two boxes containing candies are placed on a table. The boxes are labelled B_{1} and B_{2}. Box B_{1} contains 7 cinnamon candies and 4 ginger candies. Box B_{2} contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B_{1} is ^{1⁄3} and the probability of selecting box B_{2} is ^{2⁄3}. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. If he wins a colour TV, what is the probability that the marble was from the first box?

a) ^{7}⁄_{13}

b) ^{13}⁄_{7}

c) ^{7}⁄_{33}

d) ^{6}⁄_{33}

View Answer

Explanation: Let A be the event of drawing a cinnamon candy.

Let B

_{1}be the event of selecting box B

_{1}.

Let B

_{2}be the event of selecting box B

_{2}.

Then, P(B_{1}) = ^{1⁄3} and P(B_{2}) = ^{2⁄3}

Given that Suresh won the TV, the probability that the cinnamon candy was selected from B1 is

P(B_{1}|A) = (P(A|B_{1}) * P( B_{1}) ) /( P(A│B_{1} ) * P( B_{1} ) + P(A│B_{1} ) * P(B_{2}) )

= ^{7}⁄_{13}.

5. Suppose box A contains 4 red and 5 blue coins and box B contains 6 red and 3 blue coins. A coin is chosen at random from the box A and placed in box B. Finally, a coin is chosen at random from among those now in box B. What is the probability a blue coin was transferred from box A to box B given that the coin chosen from box B is red?

a) ^{15}⁄_{29}

b) ^{14}⁄_{29}

c) ^{1}⁄_{2}

d) ^{7}⁄_{10}

View Answer

Explanation: Let E represent the event of moving a blue coin from box A to box B. We want to find the probability of a blue coin which was moved from box A to box B given that the coin chosen from B was red. The probability of choosing a red coin from box A is P(R) =

^{7}⁄

_{9}and the probability of choosing a blue coin from box A is P(B) =

^{5}⁄

_{9}. If a red coin was moved from box A to box B, then box B has 7 red coins and 3 blue coins. Thus the probability of choosing a red coin from box B is

^{7}⁄

_{10}. Similarly, if a blue coin was moved from box A to box B, then the probability of choosing a red coin from box B is

^{6}⁄

_{10}.

Hence, the probability that a blue coin was transferred from box A to box B given that the coin chosen from box B is red is given by

=

^{15}⁄

_{29}.

6. An urn B_{1} contains 2 white and 3 black chips and another urn B_{2} contains 3 white and 4 black chips. One urn is selected at random and a chip is drawn from it. If the chip drawn is found black, find the probability that the urn chosen was B_{1}.

a) ^{4}⁄_{7}

b) ^{3}⁄_{7}

c) ^{20}⁄_{41}

d) ^{21}⁄_{41}

View Answer

Explanation: Let E

_{1}, E

_{2}denote the vents of selecting urns B

_{1}and B

_{2}respectively.

Then P(E

_{1}) = P(E

_{2}) =

^{1}⁄

_{2}

Let B denote the event that the chip chosen from the selected urn is black .

Then we have to find P(E

_{1}/B).

By hypothesis P(B /E

_{1}) =

^{3}⁄

_{5}

and P(B /E

_{2}) =

^{4}⁄

_{7}

By Bayes theorem P(E

_{1}/B) = (P(E1)*P(B│E1))/((P(E1) * P(B│E1)+P(E2) * P(B│E2)) )

= ((1/2) * (3/5))/((1/2) * (3/5)+(1/2)*(4/7) ) = 21/41.

7. At a certain university, 4% of men are over 6 feet tall and 1% of women are over 6 feet tall. The total student population is divided in the ratio 3:2 in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?

a) ^{2}⁄_{5}

b) ^{3}⁄_{5}

c) ^{3}⁄_{11}

d) ^{1}⁄_{100}

View Answer

Explanation: Let M be the event that student is male and F be the event that the student is female. Let T be the event that student is taller than 6 ft.

P(M) =

^{2}⁄

_{5}P(F) =

^{3}⁄

_{5}P(T|M) =

^{4}⁄

_{100}P(T|F) =

^{1}⁄

_{100}

P(F│T) = (P(T│F) * P(F))/(P(T│F) * P(F) + P(T│M) * P(M))

= ((1/100) * (3/5))/((1/100) * (3/5) + (4/100) * (2/5) )

= ^{3}⁄_{11}.

8. Previous probabilities in Bayes Theorem that are changed with help of new available information are classified as

a) independent probabilities

b) posterior probabilities

c) interior probabilities

d) dependent probabilities

View Answer

Explanation: None.

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