Mathematics

Probability Distributions

457 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
  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.