1. Total probability theorem is used in Baye’s theorem.
a) True
b) False
Explanation: Total probability theorem is used in Baye’s theorem. Baye’s theorem is given by
\(P(E_{i}|A) = \frac{P(Ei)P(A∨Ei)}{∑P (Ei)P(A∨Ei)}. \)
2. Theorem of total probability is given by P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) +…..(n terms).
a) True
b) False
Explanation: The total probability P(A) is given by the sum of the product of the probability of a particular event and the probability of A given the particular event. Hence the formula is true.
3. In badminton practice session, the probability that the player A serves properly is 0.8 and that he player B serves properly is 0.9. If there are only two players, then find the probability that it is serves properly.
a) 0.75
b) 0.85
c) 0.95
d) 0.55
Explanation: Let A be the event that the service is done properly and Ei be the event that a player is selected. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{1}{2})(\frac{8}{10}) + (\frac{1}{2})(\frac{9}{10}) \)
= 0.85.
4. The probability that person A completes all the tasks assigned is 50% and that of person B is 20%. Find the probability that all the tasks are completed.
a) 0.15
b) 0.25
c) 0.35
d) 0.45
Explanation: Let A be the event that all tasks are completed and Ei is the event that a person is selected.
Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( =(\frac{1}{2})(\frac{50}{100}) + (\frac{1}{2})(\frac{20}{100}) \)
= 0.35
5. Let there be two newly launched phones A and B. The probability that phone A has good battery life is 0.7 and the probability that phone B has good battery life is 0.8. Then find the probability that a phone has a good battery life.
a) 0.65
b) 0.75
c) 0.85
d) 0.45
Explanation: Let A be the event thtat a phome has a good battery life and Ei be the event that a phone is selected. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{1}{2})(0.7) + (\frac{1}{2})(0.8) \)
= 0.75.
6. 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
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.
7. 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) \(\frac{(6*6*6)}{(30*30*30)}\)
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).
8. 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
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.
9. 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
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
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.
10. 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
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
\(P(E|R)=\frac{P(R|E)*P(E)}{P(R)}\)
=\(\frac{(\frac{6}{10})*(\frac{5}{9})}{(\frac{7}{10})*(\frac{4}{9})+(\frac{6}{10})*(\frac{5}{9})}\)
= 15⁄29.