TGPSC MPSO / ASO · Paper II · Topic 3
TGPSC MPSO / ASO Probability questions with answers
22 questions from TGPSC ASO Paper II Mock 1 with answers and TGPSC ASO Paper II Challenge 1 with answers. Try each one, then tap Show answer.
3.1Sample space, events & classical probabilityNotesPractise this chapter · Set 1 free
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A die is thrown once. Which of the following collections of events are exhaustive?
1. {1, 2, 3} and {4, 5, 6}
2. {2, 4, 6} and {2, 3, 5}
3. {1, 2, 3, 4} and {3, 4, 5, 6}
4. {1, 2}, {3, 4} and {5, 6}
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Answer B. 1, 3 and 4 only
Collections 1, 3 and 4 cover every face. Collection 2 misses the face 1. Exhaustive events may overlap, as in 3.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on basic concepts and definitions of probability)
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A sample space is S = {a, b, c, d} with P(a) = 0.1, P(b) = 0.3 and P(c) = 2P(d). What is P(c)?
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Answer C. 0.4
P(c) + P(d) = 1 − 0.1 − 0.3 = 0.6. With P(c) = 2P(d), 3P(d) = 0.6, so P(d) = 0.2 and P(c) = 0.4.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on basic concepts of probability)
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Consider the following statements about two events A and B:
1. If A and B are independent, then A and B' (complement of B) are also independent.
2. If A and B are independent, then P(A ∩ B) = P(A) + P(B).
Which of the statements given above is/are correct?
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Answer A. 1 only
P(A ∩ B') = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A)P(B'), so A and B' are independent. For independent events P(A ∩ B) = P(A)P(B), a product, not a sum.
Source: Gupta and Kapoor, Fundamentals of Mathematical Statistics, Theory of Probability
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Assertion (A): When a fair die is thrown, the events 'getting an odd number' and 'getting a number greater than 4' are not mutually exclusive.
Reason (R): Both events contain the outcome 5.
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Answer A. Both (A) and (R) are true and (R) is the correct explanation of (A)
Both are true. The events are {1, 3, 5} and {5, 6}; they share 5, so they can happen together. R gives the reason.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on basic concepts and definitions of probability)
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Consider the following statements about the relative frequency (statistical) definition of probability:
1. It defines P(A) as the limit of the relative frequency m/n of A as the number of trials n increases indefinitely.
2. It requires the outcomes of the experiment to be equally likely.
Which of the statements given above is/are correct?
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Answer A. 1 only
The statistical (von Mises) definition is P(A) = lim (n→∞) m/n, where the trials are repeated under identical conditions. It does not need equally likely outcomes; that is a condition of the classical definition.
Source: Gupta and Kapoor, Fundamentals of Mathematical Statistics, Theory of Probability
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Two different numbers are chosen at random from 1, 2, 3, …, 9. What is the probability that their sum is even?
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Answer B. 4/9
The probability is 4/9. There are C(9, 2) = 36 equally likely pairs. The sum is even when both numbers are odd, C(5, 2) = 10 ways, or both are even, C(4, 2) = 6 ways. So the probability is 16/36 = 4/9.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
3.2Addition & multiplication theoremsNotesPractise this chapter · Set 1 free
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A and B throw a fair die alternately, A throwing first. The first to throw a six wins. What is the probability that A wins?
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Answer D. 6/11
A wins on throw 1, 3, 5, ...; these cases are mutually exclusive. Each needs earlier failures (multiplication theorem). Sum = (1/6)/(1 − 25/36) = 6/11.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability)
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Events E₁, E₂ and E₃ are mutually exclusive and exhaustive. Which of the following are necessarily true?
1. P(E₁) + P(E₂) + P(E₃) = 1
2. P(E₁ ∩ E₂) = 0
3. P(E₁) = P(E₂) = P(E₃) = 1/3
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Answer B. 1 and 2 only
Mutually exclusive and exhaustive events form a partition: their probabilities add to 1 and their pairwise intersections have probability 0. Their individual values can differ.
Source: Standard result (BRAOU Sem1 Descriptive Statistics & Probability, unit on the addition theorem)
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A box contains 10 bulbs, of which 3 are defective. Two bulbs are picked one after another without replacement. What is the probability that at least one of them is defective?
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Answer A. 8/15
P(no defective) = (7/10)(6/9) = 42/90 = 7/15. So P(at least one defective) = 1 − 7/15 = 8/15.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability)
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A die is thrown once. Which of the following pairs of events are NOT mutually exclusive?
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Answer B. 'A prime number' and 'an even number'
Primes on a die are {2, 3, 5} and evens are {2, 4, 6}; they share 2, so they are not mutually exclusive.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on events and algebra of events)
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Consider the following statements:
1. For any two events A and B, P(A ∪ B) ≥ P(A).
2. For any two events A and B, P(A ∪ B) = P(A) + P(B).
Which of the statements given above is/are correct?
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Answer A. 1 only
Statement 1 holds because A ⊂ A ∪ B. Statement 2 is the special case for mutually exclusive events only.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, units on probability)
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For two events A and B, P(A) = 1/2, P(B | A) = 1/3 and P(A | B) = 1/4. Consider the following statements:
1. P(B) = 2/3
2. A and B are independent.
3. P(A ∪ B) = 1
4. The probability that neither A nor B occurs is 1/6.
Which of the statements given above are correct?
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Answer C. 1 and 3 only
Statements 1 and 3 are correct. By the multiplication theorem, P(A ∩ B) = P(A)P(B | A) = 1/6, and then P(B) = P(A ∩ B)/P(A | B) = (1/6)/(1/4) = 2/3. P(A ∪ B) = 1/2 + 2/3 − 1/6 = 1, so the probability of neither event is 0, not 1/6. A and B are not independent, because P(A)P(B) = 1/3 differs from P(A ∩ B) = 1/6.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
3.2Conditional probability & independent eventsPractise this chapter · Set 1 free
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In a college, 60% of students study Mathematics, 40% study Statistics and 25% study both. A student chosen at random does not study Statistics. What is the probability that the student studies Mathematics?
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Answer B. 7/12
P(M | S') = P(M ∩ S')/P(S') = (0.6 − 0.25)/(1 − 0.4) = 0.35/0.6 = 7/12.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability)
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For events A and B, P(A) = 0.5, P(B) = 0.4 and P(A | B) = 0.6. What is P(B | A)?
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Answer B. 0.48
Multiplication theorem: P(A ∩ B) = P(B)P(A | B) = 0.24. Then P(B | A) = 0.24/0.5 = 0.48.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability)
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A and B are independent events. Which of the following pairs of events is NOT necessarily independent?
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Answer D. A and A ∪ B
Independence of A and B carries over to A and B', A' and B, and A' and B'. But A ⊂ A ∪ B, so knowing A occurred makes A ∪ B certain; A and A ∪ B are generally dependent.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on independent events)
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A fair die is thrown twice. Given that the first throw shows 4, what is the probability that the sum of the two throws is at least 9?
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Answer C. 1/3
Given the first throw is 4, a sum of at least 9 needs 5 or 6 on the second throw: P = 2/6 = 1/3.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability)
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For two events, P(A) = 0.3, P(B) = 0.5 and P(A | B) = 0.4. Consider the following statements:
I. P(A | B′) = 0.6, where B′ is the complement of B.
II. P(B | A) = 2/3.
Which of the statements given above is/are correct?
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Answer B. Only II is correct
Only statement II is correct. P(A ∩ B) = P(A | B)P(B) = 0.4 × 0.5 = 0.2, so P(B | A) = 0.2/0.3 = 2/3. P(A ∩ B′) = 0.3 − 0.2 = 0.1 and P(B′) = 0.5, so P(A | B′) = 0.1/0.5 = 0.2, not 0.6. The value 0.6 is 1 − P(A | B), which is P(A′ | B), a different event.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
3.2Bayes' theorem & its applicationsPractise this chapter · Set 1 free
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A disease affects 10% of a population. A test is positive for 90% of people with the disease and negative for 90% of people without it. A person tests NEGATIVE. What is the probability that the person still has the disease?
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Answer B. 1/82
P(− | D) = 0.1, P(− | D′) = 0.9. P(D ∩ −) = 0.1 × 0.1 = 0.01; P(D′ ∩ −) = 0.9 × 0.9 = 0.81. P(D | −) = 0.01/0.82 = 1/82.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on Bayes' theorem)
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A coin is picked at random from a bag holding one fair coin and one double-headed coin, and tossed. Consider the following statements:
1. If the toss gives a tail, the coin is certainly the fair coin.
2. If the toss gives a head, the coin is more likely to be the fair coin.
Which of the statements given above is/are correct?
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Answer A. 1 only
Statement 1 is true: P(tail | double-headed) = 0, so the posterior for the fair coin is 1. Statement 2 is false: P(fair | head) = (1/4)/(3/4) = 1/3, so the double-headed coin is more likely.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on Bayes' theorem)
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In a town it rains on 1 day in 5. A weather forecaster predicts rain on 80% of the days that turn out rainy and on 10% of the days that turn out dry. Rain is predicted for tomorrow. What is the probability that it will rain?
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Answer C. 2/3
P(R ∩ F) = 0.2 × 0.8 = 0.16; P(R′ ∩ F) = 0.8 × 0.1 = 0.08. P(R | F) = 0.16/0.24 = 2/3.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, unit on Bayes' theorem)
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Which of the following problems are usually solved with Bayes' theorem?
1. Finding which machine most likely produced a defective item
2. Finding the chance that a person who tested positive actually has the disease
3. Finding the arithmetic mean of a frequency distribution
4. Finding which urn a ball of a known colour came from
Select the correct answer using the code given below.
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Answer D. 1, 2 and 4 only
Statements 1, 2 and 4 all ask for the probability of a cause after an effect is seen, which is Bayes' theorem; finding a mean is not.
Source: Standard definition (BRAOU Sem1 Descriptive Statistics & Probability, unit on Bayes' theorem)
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A box has two coins: one fair and one biased that shows heads with probability 3/4. A coin is picked at random and tossed three times, giving two heads and one tail. What is the probability that the biased coin was picked?
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Answer B. 9/17
The probability is 9/17. By the binomial rule, P(2 heads in 3 | fair) = 3 × (1/2)³ = 24/64 and P(2 heads in 3 | biased) = 3 × (3/4)² × (1/4) = 27/64. With equal priors, P(biased | data) = 27/64 ÷ (27/64 + 24/64) = 27/51 = 9/17. The tail is evidence for the fair coin, so the posterior is only a little above 1/2.
Source: Standard method (BRAOU Sem1 Descriptive Statistics & Probability, probability)
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