This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Baye’s 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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