Discrete Probability Distributions Questions and Answers

Q: If a random variable \(X\) assumes only positive integral values, with the probability\[ P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}, \]\(x = 1, 2, 3, \ldots,\)then \(E(X)\) is

The expectation is 3/2.\(P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}\) is the geometric distribution counting trials up to the first success, with success probability \(p = \frac{2}{3}\). Its mean is \(1/p\):\[ \begin{aligned} E(X) &= \sum_{x \ge 1} x\cdot\tfrac{2}{3}\left(\tfrac{1}{3}\right)^{x-1} \\ &= \tfrac{2}{3}\cdot\frac{1}{(1 - 1/3)^{2}} = \tfrac{2}{3}\cdot\tfrac{9}{4} = \tfrac{3}{2} \end{aligned} \]The option 2/3 is \(p\) itself, 2/9 is \(p(1-p)\), and 1 would mean the variable always equals 1. Since \(X \ge 1\) with positive probability of larger values, its mean must exceed 1, which already rules out three options.

Q: Which of the following distribution has, mean = variance ?

The distribution with mean equal to variance is the Poisson.A Poisson distribution with parameter \(\lambda\) has mean \(\lambda\) and variance \(\lambda\), which is used as a quick check on whether count data (accidents, colonies on a plate, defects) follow it.The binomial has mean \(np\) and variance \(npq\), which is always smaller than the mean because \(q \lt 1\). In the normal distribution the mean \(\mu\) and variance \(\sigma^{2}\) are independent parameters. A chi-square distribution with \(k\) degrees of freedom has mean \(k\) and variance \(2k\).

Q: In a book of 520 pages, 390 typographical errors occur. Assuming Poisson probability law for the number of errors per page, what is the probability that a number sample of 5 pages will contain no error ?

The probability is \(e^{-3.75}\).The mean number of errors per page is \(390/520 = 0.75\), so in 5 pages the mean is \(m = 5 \times 0.75 = 3.75\). For a Poisson variable, the chance of zero events is \(e^{-m}\):\[ P(X = 0) = \frac{e^{-m} m^{0}}{0!} = e^{-3.75} \]The options with positive exponents are greater than 1 and cannot be probabilities. \(e^{-4.75}\) adds 1 to the mean by mistake.

Q: Mean is the same as variance for a :

Mean equals variance for a Poisson distribution.For a Poisson distribution with parameter \(\lambda\), both the mean and the variance equal \(\lambda\). This is often used as a quick test of whether count data are Poisson.For the binomial distribution, mean \(np\) and variance \(np(1 - p)\) differ unless \(p = 0\). A normal distribution has independent mean and variance parameters. A uniform distribution on \([a, b]\) has mean \(\frac{a+b}{2}\) and variance \(\frac{(b-a)^{2}}{12}\), equal only by coincidence.

Q: Let X follow a Poisson distribution with probability mass function P(x = 1) = P(x = 3). What is the mean value of this distribution ?

The mean is \(\sqrt{6}\), about 2.45.For a Poisson distribution with mean \(\lambda\), \(P(x) = e^{-\lambda}\lambda^{x}/x!\). Setting the probabilities of 1 and 3 equal:\[ \begin{aligned} \frac{e^{-\lambda}\lambda^{1}}{1!} &= \frac{e^{-\lambda}\lambda^{3}}{3!} \\ \lambda &= \frac{\lambda^{3}}{6} \\ \lambda^{2} &= 6,\quad \lambda = \sqrt{6} \end{aligned} \]The mean must be positive, so the negative root is rejected. \(\sqrt{2}\) and \(\sqrt{3}\) come from using 2 or 3 instead of 3! = 6 in the denominator. 7 has no basis in the working. Since a Poisson mean equals its variance, the variance is also \(\sqrt{6}\).

Q: Let X follows binomial distribution with mean and variance are 4 and \(\frac{4}{3}\). What is probability of (x ≥ 1) ?

The probability is 0.99863.For a binomial distribution, mean \(np = 4\) and variance \(npq = 4/3\). Dividing:\[ \begin{aligned} q &= \frac{4/3}{4} = \frac{1}{3} \\ p &= 1 - q = \frac{2}{3} \\ n &= \frac{4}{2/3} = 6 \end{aligned} \]\[ \begin{aligned} P(x \ge 1) &= 1 - P(x = 0) \\ &= 1 - q^{6} \\ &= 1 - \frac{1}{729} = 0.99863 \end{aligned} \]0.00137 is \(P(x = 0) = 1/729\) itself, the answer to the complementary question. 0.71256 and 0.81343 do not correspond to this distribution.

If a random variable \(X\) assumes only positive integral values, with the probability

\[ P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}, \]

\(x = 1, 2, 3, \ldots,\)

then \(E(X)\) is

A.2/9
B.2/3
C.1
D.3/2

Exam Year(s)

GATE Mathematics - 2012,

Answer: D

The expectation is 3/2.

\(P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}\) is the geometric distribution counting trials up to the first success, with success probability \(p = \frac{2}{3}\). Its mean is \(1/p\):

\[ \begin{aligned} E(X) &= \sum_{x \ge 1} x\cdot\tfrac{2}{3}\left(\tfrac{1}{3}\right)^{x-1} \\ &= \tfrac{2}{3}\cdot\frac{1}{(1 - 1/3)^{2}} = \tfrac{2}{3}\cdot\tfrac{9}{4} = \tfrac{3}{2} \end{aligned} \]

The option 2/3 is \(p\) itself, 2/9 is \(p(1-p)\), and 1 would mean the variable always equals 1. Since \(X \ge 1\) with positive probability of larger values, its mean must exceed 1, which already rules out three options.

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Which of the following distribution has, mean = variance ?

A.Binomial
B.Poisson
C.Normal
D.Chi-square

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The distribution with mean equal to variance is the Poisson.

A Poisson distribution with parameter \(\lambda\) has mean \(\lambda\) and variance \(\lambda\), which is used as a quick check on whether count data (accidents, colonies on a plate, defects) follow it.

The binomial has mean \(np\) and variance \(npq\), which is always smaller than the mean because \(q \lt 1\). In the normal distribution the mean \(\mu\) and variance \(\sigma^{2}\) are independent parameters. A chi-square distribution with \(k\) degrees of freedom has mean \(k\) and variance \(2k\).

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In a book of 520 pages, 390 typographical errors occur. Assuming Poisson probability law for the number of errors per page, what is the probability that a number sample of 5 pages will contain no error ?

A.\(e^{4.75}\)
B.\(e^{3.75}\)
C.\(e^{-4.75}\)
D.\(e^{-3.75}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The probability is \(e^{-3.75}\).

The mean number of errors per page is \(390/520 = 0.75\), so in 5 pages the mean is \(m = 5 \times 0.75 = 3.75\). For a Poisson variable, the chance of zero events is \(e^{-m}\):

\[ P(X = 0) = \frac{e^{-m} m^{0}}{0!} = e^{-3.75} \]

The options with positive exponents are greater than 1 and cannot be probabilities. \(e^{-4.75}\) adds 1 to the mean by mistake.

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Mean is the same as variance for a :

A.Poisson distribution
B.Normal distribution
C.Uniform distribution
D.Binomial distribution

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

Mean equals variance for a Poisson distribution.

For a Poisson distribution with parameter \(\lambda\), both the mean and the variance equal \(\lambda\). This is often used as a quick test of whether count data are Poisson.

For the binomial distribution, mean \(np\) and variance \(np(1 - p)\) differ unless \(p = 0\). A normal distribution has independent mean and variance parameters. A uniform distribution on \([a, b]\) has mean \(\frac{a+b}{2}\) and variance \(\frac{(b-a)^{2}}{12}\), equal only by coincidence.

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Let X follow a Poisson distribution with probability mass function P(x = 1) = P(x = 3). What is the mean value of this distribution ?

A.\(\sqrt{2}\)
B.\(\sqrt{3}\)
C.\(\sqrt{6}\)
D.7

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The mean is \(\sqrt{6}\), about 2.45.

For a Poisson distribution with mean \(\lambda\), \(P(x) = e^{-\lambda}\lambda^{x}/x!\). Setting the probabilities of 1 and 3 equal:

\[ \begin{aligned} \frac{e^{-\lambda}\lambda^{1}}{1!} &= \frac{e^{-\lambda}\lambda^{3}}{3!} \\ \lambda &= \frac{\lambda^{3}}{6} \\ \lambda^{2} &= 6,\quad \lambda = \sqrt{6} \end{aligned} \]

The mean must be positive, so the negative root is rejected. \(\sqrt{2}\) and \(\sqrt{3}\) come from using 2 or 3 instead of 3! = 6 in the denominator. 7 has no basis in the working. Since a Poisson mean equals its variance, the variance is also \(\sqrt{6}\).

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Let X follows binomial distribution with mean and variance are 4 and \(\frac{4}{3}\). What is probability of (x ≥ 1) ?

A.0.00137
B.0.71256
C.0.81343
D.0.99863

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The probability is 0.99863.

For a binomial distribution, mean \(np = 4\) and variance \(npq = 4/3\). Dividing:

\[ \begin{aligned} q &= \frac{4/3}{4} = \frac{1}{3} \\ p &= 1 - q = \frac{2}{3} \\ n &= \frac{4}{2/3} = 6 \end{aligned} \]

\[ \begin{aligned} P(x \ge 1) &= 1 - P(x = 0) \\ &= 1 - q^{6} \\ &= 1 - \frac{1}{729} = 0.99863 \end{aligned} \]

0.00137 is \(P(x = 0) = 1/729\) itself, the answer to the complementary question. 0.71256 and 0.81343 do not correspond to this distribution.

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