Category: Probability & Statistics Objective Questions

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Weibull Distribution”.

1. How many parameters are there in Weibull distribution?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: There are 3 parameters in Weibull distribution β is the shape parameter also known as the Weibull slope, η is the scale parameter, γ is the location parameter.

2. Weibull distribution gives the failure rate proportional to the power of time.
a) True
b) False
Answer: a
Clarification: The Weibull distribution function is used to get the failure rate of device proportional to the power of time. We can predict the failure of a device using Weibull distribution.

3. In Weibull distribution, if the value of β equal us equal to one that indicates the failure rate is constant over time.
a) True
b) False
Answer: a
Explanation In Weibull distribution for β < 1 failure rate that decreases with time. β = 1 have a constant failure rate. β > 1 have a failure rate that increases with time.

4. For β greater than 1 there is an inflection point for Weibull function at ___________
a) (e^1/β-1)/e^1/β-1
b) (e^1/β-1)/e^1/β
c) (e^1/β-1)/e^1/β-2
d) (e^1/β-1)/e^1/β-5
Answer: b
Clarification: There is an inflection point for Weibull function at (e^1/β-1)/e^1/β. The function is first convex, then concave after the inflection point.

5. How can we increase the height of the graph of Weibull distribution?
a) if η is decreased while β and γ are constant
b) if η is increased while β and γ are constant
c) if η is constant while β and γ are increased
d) if η is constant while β and γ are decreased
Answer: a
Clarification: If η is decreased while β and γ are kept constant, the distribution is pushed to its left and thus its height increases.

6. What is the effect of the threshold parameter γ on the graph of Weibull distribution?
a) graph shift to left or right
b) graph shifts up or down
c) height of the graph decreases
d) height of the graph increases
Answer: a
Clarification: γ is called a location parameter. The value of γ decides the least time of failure of the device. More the γ more is the graph shifted towards right.

7. What is the mean time to failure if time to failure of a gadget follows Weibull distribution with scale = 1000 hours and shape = 0.5?
a) 2500 hours
b) 1500 hours
c) 3000 hours
d) 2000 hours
Answer: d
Clarification: The failure of the electric bulb follows a Weibull Distribution,
Mean time to failure is given by
1000 × Γ (1+1/0.5)
1000 × 2 = 2000 hours.

8. The mean time of failure is given by the equation __________
a) E(X) = Γ(1/ 2β + 1)
b) E(X) = Γ(1/ β + 1)
c) E(X) = Γ(1/ β + 2)
d) E(X) = Γ(2/ β + 1)
Answer: b
Clarification: The mean time of failure is given by the equation E(X) = Γ(1/ β + 1). This represents the time expected for a device to fail.

9. The time to failure of an electric bulb follows a Weibull distribution with scale = 2000 hours and shape = 0.5. What is the probability that the electric bulb will last more than 4000 hours? What is the mean time to failure?
a) 25.3%
b) 24.3%
c) 26.3%
d) 27.3%
Answer: b
Clarification: The failure of the electric bulb follows a Weibull Distribution
The probability that the bulb will last no more than 3000 hours = WEIBULL (3000, 0.75, 1000, True) = 0.757
Probability that the bulb will last more than 4000 hours = 1 – 0.757 = 24.3%.

10. γ in Weibull distribution graph indicates _________
a) earliest time of failure
b) maximum time of failure
c) shape of the graph
d) scale of the graph
Answer: a
Clarification: γ in Weibull distribution graph indicates the location for the graph of a density function. The starting point gives the least time of a device to fail.

Probability and Statistics Assessment Questions on “Set Theory of Probability – 2”.

1. If P(^B⁄_A) = p(b), then P(A ∩ B) = ____________
a) p(b)
b) p(a)
c) p(b).p(a)
d) p(a) + p(b)
Answer: c
Clarification: P(B/A) = p(b) implies A and B are independent events
Therefore, P(A ∩ B) = p(a).p(b).

2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) ¹⁄₂
b) ¹⁄₃
c) ¹⁄₆
d) ³⁄₄
Answer: d
Clarification: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to maximum one head,
Therefore, probability = ³⁄₄.

3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then P(A/B) = ?
a) 1 – P(A /B)
b) (Pfrac{(bar{A})}{(bar{B})})
c) Not Defined
d) (frac{1-P(A cup B)}{P(bar{B}})
Answer: d
Clarification: From definition of conditional probability we have
(P(bar{A}/bar{B})=frac{bar{A}capbar{B}}{P(bar{B})})
Using De Morgan’s Law
=(frac{P(bar{A cup B})}{(bar{B}})
=(frac{1-P(A cup B)}{P(bar{B}})

4. If A and B are two events, then the probability of exactly one of them occurs is given by ____________
a) P(A ∩ B) + P(A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)
Answer: a
Clarification: The set corresponding to the required outcome is
(A ∩ B) ∪ (A ∩ B)
Hence the required probability is
P(P(A ∩ B) ∪ (A ∩ B)) = (A ∩ B) + P(A ∩ B).

5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(M) + P(N) is?
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate
Answer: b
Clarification: Given : P(M ∪ N) = 0.6, P(M ∩ N) = 0.2
P(M ∪ N) + P(M ∩ N) = P(M) + P(N)
2 – (P(M ∪ N) + P(M ∩ N)) = 2 – (P(M) + P(N))
= (1 – P(M)) + (1 – P(N))
2 – (0.6 + 0.2) = P(M) + P(N)
P(M) + P(N) = 2 – 0.8
= 1.2

6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red?
a) (frac{xy-y}{x^2+y^2+2xy-(x+y)})
b) (frac{xy}{x^2+y^2+2xy-(x+y)})
c) (frac{y^2-y}{x^2+y^2+2xy-(x+y)})
a) (frac{xy-y}{x^2+y^2+2xy-(x-y)})
Answer: b
Clarification: Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Red=(frac{x}{x+y-1})
Required porbability=(frac{y}{(x+y)}frac{x}{(x+y-1)}=frac{xy}{x^2+y^2+2xy-(x+y)})

7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is?
a) 0.29
b) 0.37
c) 0.61
d) 0.75
Answer: b
Clarification: Given: p(b) = 0.33, P(c) = 0.42
P(B ∩ C) = 0.12
P(B ∩ C) = ?
P(B ∩ C) = 1 – P(B ∪ C)
= 1 – p(b) – P(c) + P(B ∩ C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.

8. A coin is biased so that its chances of landing Head is ²⁄₃. If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is?
a) ⁴⁄₂₇
b) ⁸⁄₂₇
c) ⁴⁄₉
d) ²⁄₉
Answer: a
Clarification: Required probability = ²⁄₃ x ²⁄₃ x ¹⁄₃ = ⁴⁄₂₇.

9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) ²⁄₇
b) ⁵⁄₇
c) ⁴⁄₂₁
d) ¹⁷⁄₂₁
Answer: c
Clarification:Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (H ∩ W)
P (H ∩ W) = P (H) x P (W)
= (1 – P (H)) x (1 – P (W))
= (1 – ²⁄₃) x (1 – ³⁄₇)
= ⁴⁄₂₁.

10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then P (A|B) = ?
a) 0.5
b) 0
c) 0.25
d) 1
Answer: d
Clarification: We know that,
P (A│B) = P(A ∩ B)/P(B)
= P((A ∪ B)/P(B))
= (1 – P(A ∪ B)) /P(B)
= (1 – 0.5)/0.5
= 1.

11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) ³⁄₄
b) ¹⁄₄
c) ¹⁄₂
d) ¹⁄₆
Answer: b
Clarification:S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = ²⁄₈ = ¹⁄₄.

12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) (frac{C_2^5}{C_2^8}+frac{C_2^3}{C_2^8})
b) (frac{C_2^5 * C_2^3}{C_2^8})
c) (frac{C_1^5 * C_1^3}{C_2^8})
d) 0.5
Answer: a
Clarification:Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = ⁸C₂
Probability of picking both balls as red = ⁵C₂ /⁸C₂
Probability of picking both balls as yellow = ³C₂ /⁸C₂
∴ required probability (frac{C_2^5}{C_2^8}+frac{C_2^3}{C_2^8}).

Probability and Statistics Assessment Questions,

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Sampling Distribution – 1”.

1. What does the central limit theorem state?
a) if the sample size increases sampling distribution must approach normal distribution
b) if the sample size decreases then the sample distribution must approach normal distribution
c) if the sample size increases then the sampling distribution much approach an exponential distribution
d) if the sample size decreases then the sampling distribution much approach an exponential distribution
Answer: a
Clarification: The central limit theorem states that if the sample size increases sampling distribution must approach normal distribution. Generally a sample size more than 30 us considered as large enough.

2. Standard error is always non- negative.
a) True
b) False
Answer: a
Clarification: When we square the mean for standard deviations any negative value becomes positive. The addition of all the positive values results in a positive value. Then the square root of the positive value is also positive. Hence all standard deviations are non-negative.

3. Sampling error increases as we increase the sampling size.
a) True
b) False
Answer: b
Clarification: Sampling error is inversely proportional to the sampling size. As the sampling size increases the sampling error decreases.

4. The difference between the sample value expected and the estimates value of the parameter is called as?
a) bias
b) error
c) contradiction
d) difference
Answer: a
Clarification: The difference between the expected sample value and the estimated value of parameter is called as bias. A sample used to estimate a parameter is unbiased if the mean of its sampling distribution is exactly equal to the true value of the parameter being estimated.

5. In which of the following types of sampling the information is carried out under the opinion of an expert?
a) quota sampling
b) convenience sampling
c) purposive sampling
d) judgement sampling
Answer: d
Clarification: In judgement sampling is carried under an opinion of an expert. The judgement sampling often results in a bias because of the variance in the expert opinion.

6. Which of the following is a subset of population?
a) distribution
b) sample
c) data
d) set
Answer: b
Clarification: In sampling distribution we take a subset of population which is called as a sample. The main advantage of this sample is to reduce the variability present in the statistics.

7. The sampling error is defined as?
a) difference between population and parameter
b) difference between sample and parameter
c) difference between population and sample
d) difference between parameter and sample
Answer: c
Clarification: In sampling distribution the sampling error is defined as the difference between population and the sample. Sampling error can be reduced by increasing the sample size.

8. Any population which we want to study is referred as?
a) standard population
b) final population
c) infinite population
d) target population
Answer: d
Clarification: In sampling distribution we take a part of a population under study which is called as target population. Target population is also called as a sample.

9. Suppose we want to make a voters list for the general elections 2019 then we require __________
a) sampling error
b) random error
c) census
d) simple error
Answer: c
Clarification: Study of population is called a Census. Hence for making a voter list for the general elections 2019 we require Census.

10. Selection of a football team for FIFA World Cup is called as?
a) random sampling
b) systematic sampling
c) purposive sampling
d) cluster sampling
Answer: c
Clarification: A purposive sampling is defined as the sampling done on the basis of the characteristics of the population. Hence selecting a football team for the FIFA World cup is a purposive sampling.

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Theorem of Total Probability”.

1. If 40% of boys opted for maths and 60% of girls opted for maths, then what is the probability that maths is chosen if half of the class’s population is girls?
a) 0.5
b) 0.6
c) 0.7
d) 0.4
Answer: a
Clarification: Let E be the event of electing boy or a girl and A be the event of selecting a maths student.
P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂)
(= (frac{1}{2})(frac{40}{100}) + (frac{1}{2}) (frac{60}{100}) )
= 0.5.

2. Company A produces 10% defective products, Company B produces 20% defective products and C produces 5% defective products. If choosing a company is an equally likely event, then find the probability that the product chosen is defective.
a) 0.22
b) 0.12
c) 0.11
d) 0.21
Answer: b
Clarification: Let A be the event of selecting a defective item. Let Ei be the event of selecting a company. Then,
P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃)
( = (frac{1}{3})(frac{10}{100}) + (frac{1}{3})(frac{20}{100}) + (frac{1}{3})(frac{5}{100}) )
( = frac{0.35}{3} = 0.12. )

3. Suppose 5 men out of 100 men and 10 women out of 250 women are colour blind, then find the total probability of colour blind people. (Assume that both men and women are in equal numbers.)
a) 0.45
b) 0.045
c) 0.05
d) 0.5
Answer: b
Clarification: Let A be the event of selecting a colour blind person and Ei be the event of selecting a person. Then,
P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂)
(= (frac{1}{2})(frac{5}{100}) + (frac{1}{2})(frac{10}{250}) )
= 0.045.

4. A problem is given to 5 students P, Q, R, S, T. If the probability of solving the problem individually is 1/2, 1/3, 2/3, 1/5, 1/6 respectively, then find the probability that the problem is solved.
a) 0.47
b) 0.37
c) 0.57
d) 0.27
Answer: b
Clarification: Let A be the event that the problem is solved. Let E_i be the event that a student is chosen. Then,
P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + P(E₃) P(A|E₃) + P(E₄) P(A|E₄) + P(E₅) P(A|E₅)
( = (frac{1}{5})(frac{1}{2}) + (frac{1}{5})(frac{1}{3}) + (frac{1}{5})(frac{2}{3}) + (frac{1}{5})(frac{1}{5}) + (frac{1}{5})(frac{1}{6}) )
= 0.37.

5. The probability that the political party A does a particular work is 30% and the political party B doing the same work is 40%. Then find the probability that the work is completed if the probability of choosing the political party A is 40% and that of B is 60%.
a) 0.12
b) 0.24
c) 0.36
d) 0.48
Answer: c
Clarification: Let A be the event of completing the work and Ei be the event of selecting the political party. Then,
P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂)
( = (frac{40}{100})(frac{30}{100}) + (frac{60}{100})(frac{40}{100}) )
= 0.36.

6. Total probability theorem is used in Baye’s theorem.
a) True
b) False
Answer: a
Clarification: 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)}. )

7. 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
Answer: a
Clarification: 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.

8. 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
Answer: b
Clarification: 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.

9. 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
Answer: c
Clarification: 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

10. 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
Answer: b
Clarification: 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.

Probability and Statistics Interview Questions on “Sampling Distribution – 2”.

1. A population has N items. Samples of size n are selected without replacement. Find the number of possible samples.
a) ^NC_n
b) ⁿC_N
c) 2ⁿ
d) 2^N
Answer: a
Clarification: The number of ways of selecting and samples of size n from a population containing N atoms is ^NC_n. The probability of selecting of each sample is 1/^NC_n.

2. Find the number of all possible samples from a population containing 8 items from which 2 items are selected at random without replacement.
a) 56
b) 28
c) 66
d) 38
Answer: b
Clarification: The number of ways of selecting n samples from a population containing n items is ^NC_n. The population is N = 8 and sample size is n = 2. Therefore the number of possible samples are ⁸C₂ = 28.

3. A bag contains 6 balls of different colours. A student selects 2 balls at random without replacement. Find all possible combinations of the colours of the selected balls.
a) 13
b) 14
c) 15
d) 16
Answer: c
Clarification: Considering the experiment to be a sampling distribution where the population contains 6 balls and each sample contains 2 balls. The number of possible samples are ^NC_n that is ⁶C₂ = 15 samples.

4. Consider a population containing N items and n are selected as a sample with replacement. Find all the possible samples.
a) N
b) n^N
c) ^NC_n
d) Nⁿ
Answer: d
Clarification: The number of samples containing n items selected from a population of N items is Nⁿ. The probability of selection of each sample is 1/Nⁿ.

5. A bag contains 6 pairs of socks. If 2 pairs of socks are selected at random with replacement then the number of possible samples is?
a) 6
b) 12
c) 36
d) 216
Answer: c
Clarification: The number of samples formed with n items from a population containing N items is Nⁿ.
Here N = 6 and n = 2.
Hence samples are Nⁿ = 6² = 36.

6. Find the sampling fraction where N is population size and n is the sample size?
a) n/N
b) ^NC_n
c) n^N
d) Nⁿ
Answer: a
Clarification: In a sampling distribution if N is the population size, n is the sample size then number of sampling fractions is n/N.

7. In random sampling the probability of selecting an item from a population is unknown.
a) True
b) False
Answer: b
Clarification: A random sample is defined as the sampling in which the probability off the selecting item from a population is known. Hence it is also called as Probability Sampling.

8. In a sampling distribution the population correction factor is given by?
a) (N-1/N-n)^1/2
b) (N-n/N-1)^1/2
c) (n-1/N-n)^1/2
d) (N-1/n-1)^1/2
Answer: b
Clarification: The population correction factor is given by (N-n/N-1)^1/2. When we sample the population for more than 5% without replacement we require the population correction factor.

9. A population has 10 items and a sample has been selected from it containing 5 items. Find the finite population correction factor.
a) (5/8)^1/2
b) (5/7)^1/2
c) (5/9)^1/2
d) (5/6)^1/2
Answer: c
Clarification: The population correction factor is given by (N-n/N-1)^1/2. Here N = 10 and n = 5
(N-n/N-1)^1/2
(10-5/10-1)^1/2
(5/9)^1/2
which gives the value of correction factor as (5/9)^1/2.

10. Find the value of standard error Ẋ in a sampling distribution with replacement. Given that standard deviation of the population of 16 items is 8.
a) 3
b) 4
c) 2
d) 5
Answer: c
Clarification: Standard error in a sampling distribution with replacement is given by Ẋ = σ/(n)^1/2. Hence n = 16 and σ = 8
Ẋ = σ/(n)^1/2
Ẋ = 8/(16)^1/2
which gives the value of Ẋ = 2.

Probability and Statistics for Interviews,

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.