Probability MCQs

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Probability MCQs

श्रेणी: Mathematics | विषय: Mathematics | टॉपिक: Probability

10 प्रश्न
1
Mathematics • Probability
Consider 10 tosses of a fair die. Let X_i denote the number of times digit i appears. Find P(X_2 = 2 | X_1 = 2). Consider 10 tosses of a fair die. Let X_i denote the number of times digit i appears. Find P(X_2 = 2 | X_1 = 2).
सही उत्तर: B Correct Answer: B
Given X_1 = 2, the remaining 8 outcomes are non-one. For each such outcome, the conditional probability of getting 2 is 1/5. Hence X_2 is binomial with n = 8 and p = 1/5.
Given X_1 = 2, the remaining 8 outcomes are non-one. For each such outcome, the conditional probability of getting 2 is 1/5. Hence X_2 is binomial with n = 8 and p = 1/5.
2
Mathematics • Probability
We have P(X = x, Y = y) = 1/9 for all x, y in {1, 2, 3}. Find P(X > Y). We have P(X = x, Y = y) = 1/9 for all x, y in {1, 2, 3}. Find P(X > Y).
सही उत्तर: B Correct Answer: B
The ordered pairs satisfying X > Y are (2,1), (3,1), and (3,2). Their total probability is 3 x (1/9) = 1/3.
The ordered pairs satisfying X > Y are (2,1), (3,1), and (3,2). Their total probability is 3 x (1/9) = 1/3.
3
Mathematics • Probability
Consider 10 tosses of a biased coin with probability of obtaining a head equal to 1/3. Let Y denote the total number of tails observed. Find E(5Y + 2). Consider 10 tosses of a biased coin with probability of obtaining a head equal to 1/3. Let Y denote the total number of tails observed. Find E(5Y + 2).
सही उत्तर: C Correct Answer: C
The probability of a tail is 2/3, so E(Y) = 10 x 2/3 = 20/3. Thus E(5Y + 2) = 5E(Y) + 2 = 106/3.
The probability of a tail is 2/3, so E(Y) = 10 x 2/3 = 20/3. Thus E(5Y + 2) = 5E(Y) + 2 = 106/3.
4
Mathematics • Probability
The density of a random variable X is f_X(x) = a + bx^2 for x in [0,1], and 0 otherwise. Find (a,b) if E(X) = 3/5. The density of a random variable X is f_X(x) = a + bx^2 for x in [0,1], and 0 otherwise. Find (a,b) if E(X) = 3/5.
सही उत्तर: D Correct Answer: D
Normalization gives a + b/3 = 1, while E(X) = a/2 + b/4 = 3/5. Solving yields a = 3/5 and b = 6/5.
Normalization gives a + b/3 = 1, while E(X) = a/2 + b/4 = 3/5. Solving yields a = 3/5 and b = 6/5.
5
Mathematics • Probability
Suppose (X,Y) is jointly distributed with density f(s,t) = 1/pi if s^2 + t^2 <= 1, and 0 otherwise. Find Cov(X,Y). Suppose (X,Y) is jointly distributed with density f(s,t) = 1/pi if s^2 + t^2 <= 1, and 0 otherwise. Find Cov(X,Y).
सही उत्तर: C Correct Answer: C
By symmetry of the uniform distribution on the unit disk, E(X)=E(Y)=0 and E(XY)=0. Hence Cov(X,Y)=0.
By symmetry of the uniform distribution on the unit disk, E(X)=E(Y)=0 and E(XY)=0. Hence Cov(X,Y)=0.
6
Mathematics • Probability
Which one of the following gives a pair of independent random variables? Which one of the following gives a pair of independent random variables?
सही उत्तर: B Correct Answer: B
X depends only on the even-numbered tosses and Y only on the disjoint odd-numbered tosses. Since all tosses are independent, X and Y are independent.
X depends only on the even-numbered tosses and Y only on the disjoint odd-numbered tosses. Since all tosses are independent, X and Y are independent.
7
Mathematics • Probability
Which one of the following statements is true? Which one of the following statements is true?
सही उत्तर: C Correct Answer: C
A distribution function is right-continuous, so F(x+1/n) approaches F(x). The other statements are not true in general.
A distribution function is right-continuous, so F(x+1/n) approaches F(x). The other statements are not true in general.
8
Mathematics • Probability
What is the probability that, in a random arrangement of the English alphabet, the word MOTHER will appear? What is the probability that, in a random arrangement of the English alphabet, the word MOTHER will appear?
सही उत्तर: C Correct Answer: C
Treat MOTHER as one fixed block. Together with the other 20 letters there are 21 objects, giving 21! favourable arrangements out of 26! total arrangements.
Treat MOTHER as one fixed block. Together with the other 20 letters there are 21 objects, giving 21! favourable arrangements out of 26! total arrangements.
9
Mathematics • Probability
Let X and Y be two independent N(0,1) random variables. Find Cov(X+3Y, 5X-2Y). Let X and Y be two independent N(0,1) random variables. Find Cov(X+3Y, 5X-2Y).
सही उत्तर: B Correct Answer: B
Using independence and unit variances, Cov(X+3Y,5X-2Y)=5Var(X)-6Var(Y)=5-6=-1.
Using independence and unit variances, Cov(X+3Y,5X-2Y)=5Var(X)-6Var(Y)=5-6=-1.
10
Mathematics • Probability
Which one of the following statements is false? Which one of the following statements is false?
सही उत्तर: B Correct Answer: B
A probability density can exceed 1 on a sufficiently short interval; only its integral over the whole line must equal 1. Therefore option B is false.
A probability density can exceed 1 on a sufficiently short interval; only its integral over the whole line must equal 1. Therefore option B is false.

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