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 general knowledge math & puzzles
  1. 30.0%.

  2. 53.3%

  3. 64.0%

  4. 16.0%

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The conditional probability formula states P(A|B) = P(A ∩ B) / P(B). Given P(A ∩ B) = 0.32 and P(B) = 0.50, we calculate P(A|B) = 0.32 / 0.50 = 0.64 = 64%. This means there's a 64% chance that event A occurs when we know event B has occurred. Option D would be the joint probability P(A ∩ B), not the conditional probability.

Multiple choice technology testing
  1. Random

  2. Acceptance

  3. 100%

  4. Binomial

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The binomial distribution model applies to random sampling where each item has the same probability of being selected, and selections are independent. Acceptance sampling is a specific application, not the sampling type itself. 100% inspection is not sampling at all. Binomial sampling is redundant. Random sampling is the fundamental method that enables binomial distribution assumptions.

Multiple choice
  1. $f_1(t)+f_2(t)$
  2. $\int_0^t f_1(x)f_2(x)dx$
  3. $\int_0^t f_1(x)f_2(t-x)dx$
  4. $\max\{f_1(t),f_2(t)\}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability density function of f1 (t) and f2 (t) is = $\int_0^t f_1(x)f_2(t-x)$

Multiple choice
  1. $\frac{^{2n}\mathrm{C}_n}{4^n}$
  2. $\frac{^{2n}\mathrm{C}_n}{2^n}$
  3. $\frac{1}{^{2n}\mathrm{C}_n}$
  4. $\frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Required favourable cases = Number of ways of selecting n element out of the set of size 2n = $^{2n}\mathrm{C}_n$ As each toss may result in either head or tail, so for 2n tosses, total no. of combinations = 22n = 4n $\therefore$Required probability = $\dfrac{\text{Favourable cases}}{\text{Total cases}}$ $ = \frac{^{2n}\mathrm{C}_n}{4^n}$

Multiple choice
  1. 0.24

  2. 0.36

  3. 0.4

  4. 0.6

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The required probability = 0.4 x 0.4 + 0.6 x 0.4 = 0.16 + 0.24 = 0.40

Multiple choice
  1. pq + (1 − p) (1 − q)

  2. (1 − q)p

  3. (1 − p) q

  4. pq

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A computer is declared faulty in two cases: (1) It is actually faulty AND the test correctly identifies it: probability = p × q. (2) It is NOT faulty BUT the test wrongly declares it faulty: probability = (1-p) × (1-q). These are mutually exclusive events, so we add them: P(declared faulty) = pq + (1-p)(1-q). This is Option A. Option D (pq) only counts true positives, missing false positives. Options B and C only capture one scenario each.