Companies B1, B2 and B3 produce 30%, 45% and 25% of the cars respectively. It is known that 2%, 3% and 2% of these cars produced from B1, B2 and B3 are defective. If a car purchased is found to be defective, what is the probability that this is purchased from company B2?

Let E be the event that a car is defective. Let B1, B2, and B3 be the events that a car is produc...

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD MCA 2022 PROBABILITY AND STATISTICS Previous Question Paper (October/November - 2022 - R13)

MCA | 2022 | I | R13 | October/November - 2022

Answer PaperPaper Code: 811ADMaximum Marks: 60Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

112 marks

Companies B₁, B₂ and B₃ produce 30%, 45% and 25% of the cars respectively. It is known that 2%, 3% and 2% of these cars produced from B₁, B₂ and B₃ are defective. If a car purchased is found to be defective, what is the probability that this is purchased from company B₂?

6 marks

Answer

Let E be the event that a car is defective. Let B₁, B₂, and B₃ be the events that a car is produced by companies B₁, B₂, and B₃ respectively. The given probabilities are: P(B₁) = 0.30, P(B₂) = 0.45, P(B₃) = 0.25. The conditional probabilities of a defective car are: P(E|B₁) = 0.02, P(E|B₂) = 0.03, P(E|B₃) = 0.02.

Using the law of total probability, the total probability of a defective car is:

P(E) = P(B₁)P(E|B₁) + P(B₂)P(E|B₂) + P(B₃)P(E|B₃)

P(E) = (0.30)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.006 + 0.0135 + 0.005 = 0.0245

By Bayes' theorem, the probability that the defective car was produced by B₂ is:

P(B₂|E) = P(B₂)P(E|B₂) / P(E)

P(B₂|E) = 0.0135 / 0.0245 = 135 / 245 = 27 / 49 ≈ 0.551

For any three events A, B and C defined on the sample space S verify that

P(A∪B|C) = P(A|C) + P(B|C) - P(A∩B|C)

6 marks

Answer

By the definition of conditional probability, for any event X, P(X|C) = P(X ∩ C) / P(C), provided P(C) > 0. Applying this to the expression P(A ∪ B | C):

P(A ∪ B | C) = P((A ∪ B) ∩ C) / P(C)

Using the distributive law of sets, (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C). Thus:

P(A ∪ B | C) = P((A ∩ C) ∪ (B ∩ C)) / P(C)

Applying the addition theorem for probability, P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y), where X = A ∩ C and Y = B ∩ C:

P(A ∪ B | C) = [P(A ∩ C) + P(B ∩ C) - P((A ∩ C) ∩ (B ∩ C))] / P(C)

P(A ∪ B | C) = [P(A ∩ C) + P(B ∩ C) - P(A ∩ B ∩ C)] / P(C)

Distributing the denominator P(C):

P(A ∪ B | C) = P(A ∩ C)/P(C) + P(B ∩ C)/P(C) - P(A ∩ B ∩ C)/P(C)

By the definition of conditional probability, this simplifies to:

P(A ∪ B | C) = P(A|C) + P(B|C) - P(A ∩ B | C)

212 marks

State and Prove Baye’s Theorem.

6 marks

Answer

If A and B are independent events of a sample space S, then show that (i) A and B are independent (ii) A and B are independent (iii) A and B are independent.

6 marks

Answer

312 marks

A random variable X has the following probability function:

x	0	1	2	3	4	5	6	7
P(X=x)	0	k	2k	2k	3k	k²	2k²	7k²+k

Find (i) mean of X. (ii) variance of X.

6 marks

Answer

If two cards are drawn from a pack of 52 cards which are diamonds, using Poisson distribution find the probability of getting two diamonds at least 3 times in 51 consecutive trials of two cards drawing each time.

6 marks

Answer

412 marks

f(x) = (1/2)x e^-x

, when x ≥ 0 can be regarded as a probability function for a continuous random variable? If so, find Mean and Variance of the random variable.

6 marks

Answer

A random sample of size 64 is taken from a normal population with mean 51.4 and S.D 68. What is the Probability that the mean of the sample will (i) fall between 50.5 and 52.3 (ii) be less than 50.6.

6 marks

Answer

A finite population consists of the numbers 4, 8, 12, 16, 20, 24.

List of all possible samples of size 3 without replacement.
Find the mean of the population.
Find the standard deviation of the population.
Calculate the mean of the sampling distribution of means.
Find the standard deviation of sampling distributions of means.

12 marks

Answer

A professor’s feeling about the mean mark in the final examination in probability of a large group of students expressed subjectively by normal distribution with μ₀ = 67.2 and σ₀ = 1.5.

If the mean mark lies in the interval (65, 70) determine the prior probability the professor should assign to the mean mark.
Find the posterior mean μ₁ and standard deviation σ₁ if the examinations are conducted on a random sample of 40 students yielding mean 74.9 and S.D 7.4.
Determine the posterior probability which he will thus assign to mean mark being in the interval (65, 70) using results obtained in (b).
Construct a 95% Bayesian interval for μ.

12 marks

Answer

Two horses A and B were tested according to the time to run a particular track with the following results:

Horse A: 28, 30, 32, 33, 33, 29, 34

Horse B: 29, 30, 30, 24, 27, 29

Test whether the two horses have the same running capacity.

12 marks

Answer

A sample of 12 fathers and their eldest sons gave the following data about their height in inches. Calculate the Coefficient of the rank correlation.

Fathers	65	63	67	64	68	62	70	66	68	67	69	71
Sons	68	66	68	65	69	66	68	65	71	67	68	70

12 marks

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD MCA 2022 PROBABILITY AND STATISTICS Previous Question Paper (October/November - 2022 - R13)

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