Probability and Statistics Multiple Choice Questions & Answers (MCQs) 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
Answer: b
Clarification: 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?
a) (frac{(6*5*4)}{(30*30*30)})
b) (frac{(6*5*4)}{(30*29*28)})
c) (frac{(6*5*3)}{(30*29*28)})
d) (frac{(6*6*6)}{(30*30*30)})
Answer: b
Clarification: 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 B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 1⁄3 and the probability of selecting box B2 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
Answer: c
Clarification: 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 B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 1⁄3 and the probability of selecting box B2 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
Answer: a
Clarification: 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
Given that Suresh won the TV, the probability that the cinnamon candy was selected from B1 is
P(B1|A) = (P(A|B1) * P( B1))/( P(A│B1) * P( B1) + P(A│B1) * P(B2))
(frac{(frac{7}{11})*(frac{1}{3})}{(frac{7}{11})*(frac{1}{3})+(frac{3}{11})*(frac{2}{3})})
= 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
Answer: a
Clarification: 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
(P(E|R)=frac{P(R|E)*P(E)}{P(R)})
=(frac{(frac{6}{10})*(frac{5}{9})}{(frac{7}{10})*(frac{4}{9})+(frac{6}{10})*(frac{5}{9})})
= 15⁄29.
6. An urn B1 contains 2 white and 3 black chips and another urn B2 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 B1.
a) 4⁄7
b) 3⁄7
c) 20⁄41
d) 21⁄41
Answer: d
Clarification: Let E1, E2 denote the vents of selecting urns B1 and B2 respectively.
Then P(E1) = P(E2) = 1⁄2
Let B denote the event that the chip chosen from the selected urn is black .
Then we have to find P(E1 /B).
By hypothesis P(B /E1) = 3⁄5
and P(B /E2) = 4⁄7
By Bayes theorem P(E1 /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
Answer: c
Clarification: 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
Answer: b
Clarification: None.