Mathematics

Probability Distributions

488 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

Multiple choice

In a simple random sample, each member of the population has an:

  1. Equal chance of being selected

  2. Unequal chance of being selected

  3. No chance of being selected

  4. Chance of being selected that depends on their characteristics
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Simple random sampling ensures that every member of the population has an equal opportunity to be included in the sample, eliminating bias and increasing the representativeness of the sample.

Multiple choice

What is the probability of getting exactly 5 successes in 10 independent trials, each with a probability of success of 0.4?

  1. 0.2461

  2. 0.3025

  3. 0.4096

  4. 0.5120
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of getting exactly 5 successes in 10 independent trials, each with a probability of success of 0.4, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 5, p = 0.4, and q = 0.6. Plugging these values into the formula, we get P(5) = (10C5) * 0.4^5 * 0.6^5 = 0.2461.

Multiple choice

What is the probability of getting at least 3 successes in 10 independent trials, each with a probability of success of 0.3?

  1. 0.7461

  2. 0.6723

  3. 0.5905

  4. 0.5120
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of getting at least 3 successes in 10 independent trials, each with a probability of success of 0.3, can be calculated by subtracting the probability of getting 0, 1, or 2 successes from 1. The probability of getting 0 successes is (0.7)^10 = 0.0282, the probability of getting 1 success is 10 * (0.7)^9 * (0.3)^1 = 0.1323, and the probability of getting 2 successes is 45 * (0.7)^8 * (0.3)^2 = 0.0856. Therefore, the probability of getting at least 3 successes is 1 - 0.0282 - 0.1323 - 0.0856 = 0.7461.

Multiple choice

What is the probability of getting exactly 4 successes in 10 independent trials, each with a probability of success of 0.2?

  1. 0.1175

  2. 0.2048

  3. 0.2461

  4. 0.3025
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of getting exactly 4 successes in 10 independent trials, each with a probability of success of 0.2, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 4, p = 0.2, and q = 0.8. Plugging these values into the formula, we get P(4) = (10C4) * 0.2^4 * 0.8^6 = 0.2048.

Multiple choice

What is the probability of getting at most 2 successes in 5 independent trials, each with a probability of success of 0.6?

  1. 0.6656

  2. 0.7736

  3. 0.8438

  4. 0.9144
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The probability of getting at most 2 successes in 5 independent trials, each with a probability of success of 0.6, can be calculated by adding the probabilities of getting 0, 1, and 2 successes. The probability of getting 0 successes is (0.4)^5 = 0.01024, the probability of getting 1 success is 5 * (0.4)^4 * (0.6)^1 = 0.12288, and the probability of getting 2 successes is 10 * (0.4)^3 * (0.6)^2 = 0.28224. Therefore, the probability of getting at most 2 successes is 0.01024 + 0.12288 + 0.28224 = 0.8438.

Multiple choice

What is the probability of getting at least 3 successes in 5 independent trials, each with a probability of success of 0.7?

  1. 0.9688

  2. 0.8750

  3. 0.7578

  4. 0.6250
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of getting at least 3 successes in 5 independent trials, each with a probability of success of 0.7, can be calculated by subtracting the probability of getting 0, 1, or 2 successes from 1. The probability of getting 0 successes is (0.3)^5 = 0.00243, the probability of getting 1 success is 5 * (0.3)^4 * (0.7)^1 = 0.02187, and the probability of getting 2 successes is 10 * (0.3)^3 * (0.7)^2 = 0.07569. Therefore, the probability of getting at least 3 successes is 1 - 0.00243 - 0.02187 - 0.07569 = 0.9688.

Multiple choice

What is the probability of getting exactly 2 successes in 5 independent trials, each with a probability of success of 0.4?

  1. 0.3277

  2. 0.4096

  3. 0.5120

  4. 0.6400
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of getting exactly 2 successes in 5 independent trials, each with a probability of success of 0.4, can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 5, x = 2, p = 0.4, and q = 0.6. Plugging these values into the formula, we get P(2) = (5C2) * 0.4^2 * 0.6^3 = 0.4096.

Multiple choice

What is the probability of getting at most 1 success in 4 independent trials, each with a probability of success of 0.5?

  1. 0.3125

  2. 0.4375

  3. 0.5625

  4. 0.6875
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The probability of getting at most 1 success in 4 independent trials, each with a probability of success of 0.5, can be calculated by adding the probabilities of getting 0 and 1 successes. The probability of getting 0 successes is (0.5)^4 = 0.0625, and the probability of getting 1 success is 4 * (0.5)^3 * (0.5)^1 = 0.25. Therefore, the probability of getting at most 1 success is 0.0625 + 0.25 = 0.6875.

Multiple choice

What is the probability of getting at least 2 successes in 4 independent trials, each with a probability of success of 0.3?

  1. 0.4219

  2. 0.5063

  3. 0.5905

  4. 0.6723
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The probability of getting at least 2 successes in 4 independent trials, each with a probability of success of 0.3, can be calculated by subtracting the probability of getting 0 or 1 success from 1. The probability of getting 0 successes is (0.7)^4 = 0.2401, and the probability of getting 1 success is 4 * (0.7)^3 * (0.3)^1 = 0.3430. Therefore, the probability of getting at least 2 successes is 1 - 0.2401 - 0.3430 = 0.6723.

Multiple choice

What is the probability of an event occurring?

  1. The likelihood of an event occurring

  2. The frequency of an event occurring

  3. The impact of an event occurring

  4. The severity of an event occurring
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probability is the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates that the event is impossible and 1 indicates that the event is certain.

Multiple choice

What is the expected value of a fair Martingale?

  1. Positive

  2. Negative

  3. Zero

  4. Depends on the initial bet
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expected value of a fair Martingale is always zero, regardless of the initial bet. This is because the gains and losses over time balance out.

Multiple choice

Which of the following is an example of a Martingale?

  1. Simple random walk

  2. Geometric Brownian motion

  3. Poisson process

  4. Exponential distribution
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Geometric Brownian motion is an example of a continuous-time Martingale, commonly used in financial modeling.

Multiple choice

What is the expected value of a random variable X that takes on the values 1, 2, and 3 with probabilities 0.2, 0.5, and 0.3, respectively?

  1. 1.5

  2. 1.8

  3. 2.0

  4. 2.2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expected value of a random variable is calculated by multiplying each possible value by its probability and then summing the results. In this case, we have: E(X) = (1 * 0.2) + (2 * 0.5) + (3 * 0.3) = 0.2 + 1.0 + 0.9 = 2.0.

Multiple choice

What is the name of the mathematical theorem that states that the expected value of a random variable is equal to the sum of the products of each possible outcome and its probability?

  1. Central Limit Theorem

  2. Law of Large Numbers

  3. Bayes' Theorem

  4. Markov's Inequality
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The Expected Value Theorem, also known as the Mean Value Theorem for Random Variables, states that the expected value of a random variable is equal to the sum of the products of each possible outcome and its probability.

Multiple choice

What is the expected value of a random variable X with probability mass function P(X = x) = 1/3 for x = 1, 2, and 3?

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The expected value of a random variable is calculated as E(X) = ΣxP(X = x). In this case, E(X) = 1(1/3) + 2(1/3) + 3(1/3) = 2.