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
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.
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Random
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Acceptance
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100%
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Binomial
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.
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1/4,1/2
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1/2,1/4
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1/2,1
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1,1/2
D
Correct answer
Explanation
$ P(A/B) = \dfrac{P(A)P(B/A)}{ \sum P(A)P(B/A)} $
So, value of P(A/B) = P(A) = 1
P(B/A) = P(B) = 1/2
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1/4,1/2
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1/2,1/4
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1/2,1
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1,1/2
D
Correct answer
Explanation
$ P(A/B) = \dfrac{P(A)P(B/A)}{ \sum P(A)P(B/A)} $
So, value of P(A/B) = P(A) = 1
P(B/A) = P(B) = 1/2
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$f_1(t)+f_2(t)$
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$\int_0^t f_1(x)f_2(x)dx$
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$\int_0^t f_1(x)f_2(t-x)dx$
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$\max\{f_1(t),f_2(t)\}$
C
Correct answer
Explanation
Probability density function of f1 (t) and f2 (t) is
= $\int_0^t f_1(x)f_2(t-x)$
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$\dfrac{1}{2}$
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2
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$\sqrt{n}$
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n
B
Correct answer
Explanation
As the sample S $\subseteq $ X, each element of S is an integral power of 2. The expected value of the smallest number is sample S is 2.
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$\frac{^{2n}\mathrm{C}_n}{4^n}$
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$\frac{^{2n}\mathrm{C}_n}{2^n}$
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$\frac{1}{^{2n}\mathrm{C}_n}$
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$\frac{1}{2}$
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}$
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$\dfrac{1}{2}$
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$\dfrac{1}{10}$
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$\dfrac{9!}{20!}$
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None of these
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np (1-p)n-1
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(1- p)n-1
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p (1- p)n-1
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1- (1- p)n-1
Suppose the letters a, b, c, d, e, f have probabilities $ \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},\dfrac{1}{32},\dfrac{1}{32}$ respectively.
What is the average length of the correct answer to Q.?
C
Correct answer
Explanation
The required probability
= 0.4 x 0.4 + 0.6 x 0.4
= 0.16 + 0.24
= 0.40
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pq + (1 − p) (1 − q)
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(1 − q)p
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(1 − p) q
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pq
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.
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0 and 0.5
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0 and 1
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0.5 and 1
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0.25 and 0.75
C
Correct answer
Explanation
The cumulative distribution function
F(x) = P(X $\le$ x)
F(-1) = P(X $\le$ -1) = P(X = -1) = 0.5
F(+1) = P(X $\le$ +1) = P(X = -1)+P(x = + 1) = 0.5 + 0.5 = 1
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8(2e3)
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9/(2e3)
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17/(2e3)
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26/(2e3)
D
Correct answer
Explanation
The number of 2-digit multiples of 7 = 13
$\therefore$ Probability of choosing a number not divisible by $7 = \dfrac{90-13}{90} = \dfrac{77}{90}$