Let p be the probability of happening of an event.
Given p = 0.002 = \(\frac { 2 }{ 1000 } \)
Also n = 1000
∴ Mean = np = \(1000\times \frac { 2 }{ 1000 } \) = 2
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P( event happens exactly twice)
= P(X = 2)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -2 }.{ (2 }^{ 2 }) }{ 2 } \)
= e^-2(2) = 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Let p be the probability of winning the prize
Given p = \(\\ \frac { 1 }{ 100 } \) and n = 50
∴ Mean = np = \(\frac { 1 }{ 100 } \times 50=\frac { 1 }{ 2 } \)
∴ λ = 0.5
Hence, X follows Poisson distribution with
P(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
∴ P(winning the prize at least once) = P(X ≥ 1)
= 1-P(X < 1)
= 1-P(X = 0)
= 1-\(\frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \) = 1-e^-0.5
= 1-0.6066
∴ P(X≥1) = 0.3934

The probability function for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞ Given f(x) = \(C{ e }^{ -\left( \frac { x-100 }{ 50 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 25 } \right) ^{ 2 } }\)
= \(C{ e }^{ -\frac { 1 }{ 2 } \left( \frac { x-100 }{ 5 } \right) ^{ 2 } }\) ...(2)
Comparing (1) and (2), μ =100, σ = 5 and
C = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ 5\sqrt { 2\pi } } \)
∴ C = \(\frac { 1 }{ 5\sqrt { 2\pi } } \).

Given n = 10
Probability of selecting a defective pen = p
= \(\frac { 10 }{ 50 } =\frac { 1 }{ 5 } \)
q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
∴ P(X = x) = ⁿC_x p^xq^n-x
P (at least one pen is defective)
= P(X≥1) = 1-P(X<1)
= 1-P(X = 0)
= 1-10C₀ \(\left( \frac { 1 }{ 5 } \right) ^{ 0 }\left( \frac { 4 }{ 5 } \right) ^{ 10 }\)
= 1-\(\frac { { 4 }^{ 10 } }{ { 5 }^{ 10 } } \)

Given mean μ = 60 and S.D. σ = 5

To find P(X > 60)
When X = 60, Z =\(\frac { X-\mu }{ \sigma } =\frac { 60-60 }{ 5 } \) = 0
∴ P(X > 60) = P(Z > 0) = P (0 < Z < ∞)
= 0.5
∴ 50% of students scored more than 60 marks

Given n = 10, P(H) = \(\frac { 1 }{ 2 } \) ⇒ p =\(\frac { 1 }{ 2 } \)
q=1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
P(X = x) = nC_x p^xq^n-x
∴ P(X = 5) = 10C₅ p⁵q⁵
= \(\frac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } \left( \frac { 1 }{ 2 } \right) ^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 10 }\)
= \(\frac { 6\times 7\times 6 }{ 2^{ 10 } } \)
= \(\frac { 2\times 3\times 7\times 2\times 3 }{ 2^{ 10 } } =\frac { 63 }{ { 2 }^{ 8 } } \)
= \(\frac { 63 }{ 256 } \).

Since X is a binomial variate X ~ B (5, p)
n = 5 and P(X = x) = nC_x p^x q^n-x
Given P[X = 2] = P[X = 3]
⇒ 5C₂ p²q³ = 5C₃ p³q²

⇒ q=p
we know p+q = 1 ⇒ p+p =1
⇒ 2p = 1⇒ p =\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \).

Given mean 20 ⇒ np = 20
S.D = 4 ⇒ \(\sqrt { npq } \) = 4
∴ \(\frac { npq }{ np } =\frac { 16 }{ 20 } \Rightarrow q=\frac { 4 }{ 5 } \)
P = 1-q =\(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substitutingp and q in npq = 16, we get
\(n\times \frac { 1 }{ 5 } \times \frac { 4 }{ 5 } \) =16
n = \(\frac { 16\times 5\times 5 }{ 4 } \)  = 100
∴ Number of events = 100

Given n = 9 and mean = 6 ⇒ np = 6
9p = 6 ⇒ \(\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
∴ q=1-p = \(1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
Variance = npq = \(9\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } \) = 2

Let λ be the parameter
Given 3. P(X = 2) = P(X = 4)
⇒ 3. \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =\frac { e^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \)
\(\frac { 3{ \lambda }^{ 2 } }{ 2 } =\frac { { \lambda }^{ 4 } }{ 4\times 3\times 2 } \)
⇒ 36λ² = λ⁴
λ⁴-36λ² = 0
⇒ λ²(λ²-36) = 0
⇒ λ² = 0 or λ² = 36
⇒ λ = 6 since λ > 0

A random variable Z = \(\frac { X-\mu }{ \sigma } \) follows the standard normal distribution is called the standard normal variate with mean 0 and standard deviation 1. i.e. Z~ N(0,1). Its,probability density function is given by:
\(\varphi(Z)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{Z^{2}}{2}},-\infty< Z< \infty \)

Poisson distribution is a limiting case of binomial distribution under the following conditions.
(i) n, the number of trials is indefinitely large ie., n⟶∞.
(ii) p, the constant probability of success in each trial is very small ie., p ⟶0.
(iii) np = λ is finite.Thus p = λ/n and q = 1 -(λ/n) where λ is a positive real number.

Let p he the probability of getting a girls
∴ p = \(\frac { 1 }{ 2 } \) [∵ one favourable event and total no of events is 2]
⇒ q = 1-p = 1-\(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \) and n = 3
∴ (getting exactly 2 girls)
= P(X = 2)
= ³C₂\(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\)   [∵ p(x) = ⁿC_x p^xq^n-x, n = 3 and x = 2]
= \(3\left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \frac { 1 }{ 8 } \) = 0.375
P (getting exactly 2 girls) = 0.375.

Let p be the probability of a patient to recover
Given q = 75% = \(\frac { 75 }{ 100 } \)  = 0.75
∴ P = 1 - q = 1 - 0.75 = 0.25
n = 6
P (4 will recover) = P(X = 4)
= 6C₄ (0.25)⁴ (0.75)²
[∵ P(x) = nC_x p^xq^n-x n = 6, x = 4
= 6C₂ (0.25)⁴ (0.75)²

= 15 (0.25)⁴ (0.75)²
P(X = 4) = 0.03295

Given mean of the binomial distribution is 4.
⇒ np =4  ...(1)
Variance=3 ⇒ npq =3..(2)
(2) \(\div \) (1) given, \(\frac { npq }{ np } =\frac { 3 }{ 4 } \Rightarrow q=\frac { 3 }{ 4 } \)
∴ p=1-q =\(1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
Substituting p=\(\frac { 1 }{ 4 } \) in (1) we get,
n\(\left( \frac { 1 }{ 4 } \right) \) =4 ⇒ n=16
∴ The binomial distribution is ⁿC_x p^xq^n-x, x=0,1,2....n
⇒ ¹⁶C_x \(\left( \frac { 1 }{ 4 } \right) ^{ x }\left( \frac { 3 }{ 4 } \right) ^{ 16-x }\) x=0,1,2,......16.
∴ P(X=15) =¹⁶C₁₅ \(\left( \frac { 1 }{ 4 } \right) ^{ 15 }\left( \frac { 3 }{ 4 } \right) ^{ 1 }\)
=¹⁶C₁ \(\frac { 1 }{ { 4 }^{ 15 } } .\frac { 3 }{ 4 } =16.\frac { (3) }{ 4^{ 16 } } =\frac { 4^{ 2 }.(3) }{ 4^{ 16 } } \)
P(X=15) =\(\frac { 3 }{ { 4 }^{ 14 } } \).