Normal distribution is a limiting case of Binomial distribution under the following conditions.
(i) n, the number of trials is infinitely large i.e. n ⟶ ∞
(ii) Neither p(or q) is very small.

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:
\(\varphi(Z)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{Z^{2}}{2}},-\infty< Z< \infty \)

A random variable X is said to follow a normal distribution with parameters mean μ and variance σ², if its probability density function is given by

Poisson distribution is the only distribution in which the mean and variance are equal.

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.

(i) Number of lightnings per second.
(ii) Number of printing mistakes per page in a textbook.

A random variable X is said to follow a Poisson distribution with parameter λ if it assumes only non-negative values and it probability mass function is given by
P(x,λ) = P(X = x)  {\(=\begin{cases} \begin{matrix} \frac { e^{-λ}{ \lambda }^{ x } }{ x! } , & x=0,1,2,....,\lambda >0 \\ 0, & otherwise \end{matrix} \end{cases}\)

Given mean of a binomial distribution is 5
np = 5 ....(1)
Also, standard deviation is 2 ⇒ Variance = 2² = 4
∴ npq = 4...(2)
(2) \(\div \) (1) given, \(\frac { npq }{ np } \)=\(\frac { 4 }{ 5 } \)
⇒ q = \(\frac { 4 }{ 5 } \)
p = 1-q = \(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substituting p = \(\frac { 1 }{ 5 } \) in (1) we get
n\(\left( \frac { 1 }{ 5 } \right) \) = 5 ⇒ n = 25
∴ The binomial distribution is ⁿC_x p^xq^n-x, x = 0,1,2,....n
⇒ ²⁵C_x \(\left( \frac { 1 }{ 5 } \right) ^{ x }\left( \frac { 4 }{ 5 } \right) ^{ 25-x }\), x = 0,1,2,....25.

Let p be the probability of getting doublet in a pair of dice.
∴ p = \(\frac { 6 }{ 36 } \) [∵ favourable events are (1, 1) (2,2) (3,3) (4,4) (5,5) (6,6) and n(S) = 36]
⇒ p = \(\frac { 1 }{ 6 } \) ∴ q=1-p =\(1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)]
∴ P (getting 2 success) = P(X = 2)
=⁴C₂ \(\left( \frac { 1 }{ 6 } \right) ^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 2 }\)     [∵ p(x) =ⁿC_x p^xq^n-x, n = 4, x = 2 ]

∴ P(X = 2) =\(\frac { 25 }{ 216 } \).

Given probability of success p = 0.09
∴ q = 1 - p = 1 - 0.09 = 0.91
n = 1
Also P(atleast one success) = \(\frac { 1 }{ 3 } \) or more
∴ P(X≥1) = \(\frac { 1 }{ 3 } \)
⇒ 1-P(X < 1) = \(\frac { 1 }{ 3 } \)
⇒ P(X<1) =\(1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
⇒ P(X = 0) =\(\frac { 2 }{ 3 } \)
⇒ ⁿC_x p^xq^n-x = \(\frac { 2 }{ 3 } \)
Putting x = 0,
ⁿC₀ (0.09)⁰ (0.91)^n-0 = \(\frac { 2 }{ 3 } \)
⇒ (0.91)ⁿ = \(\frac { 2 }{ 3 } \) =0.6666
when (0.91) is Jultiplied 5 times we are getting 0.6240
∴ n = 5 or more
Here number of trails are 5 or more.

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.

The binomial distribution can be used under the following conditions.
(i) The number of trials en' is finite,
(ii) The trials are independent of each other.
(iii) The probability of success 'p' is constant for each trial.
(iv) In every trial there are only two possible outcomes namely success or failure.

A random experiment whose outcomes are of two types namely success S and failure P, occurring with probabilities p and q is called a Bernoulli trial.

A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function is given by 
P(X = x) = P(x) = q = 1-p \(=\begin{cases} \begin{matrix} { { n }_{ C } }_{ x }P^{ x }{ q }^{ n-x },x=0,1,2,...n; \end{matrix} \\ \begin{matrix} 0, & otherwise \end{matrix} \end{cases}\)

The average number of typographical errors per page in the book is given by \(\lambda\) = (390/520) = 0.75.
Hence using Poisson probability law, the probability of x errors per page is given by
\(P(X=x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } ={ e }^{ -0.75 }=\frac{(0.75)^x}{x!}\) x = 0,1,2,3……
The required probability that a random sample of 5 pages will contain no error is given by :
[P(X = 0)]⁵ = (e-0.75)⁵ = e^-3.75

Given that p(0) = 0.2725
\(\frac { { e }^{ -\lambda }{ \lambda }^{ 0 } }{ 0! } =0.2725\)
⇒ e^-\(\lambda\) = 0.2725 (by using exponent table)
\(\lambda\) = 1.3
∴ p(X=1) = e^-1.3(1.3)/1!
= e^-1.3(1.3)
= 0.2725 x 1.3
= 0.3543

Let X be a random variable follows binomial distribution with p = q = 1/2
P (3 heads) = \(5{ C }_{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ x }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-x }\)
\(={ 5C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5-3 }\)
\(=5{ C }_{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
\(=\frac { 5 }{ 16 } \)

Mean: np = 12
\(SD=\sqrt { npq } =4\)
\(npq={ 4 }^{ 2 }=16,\frac { np }{ npq } =\frac { 12 }{ 16 } =\frac { 3 }{ 4 } \)
\(q=\frac { 4 }{ 3 } >1\)
Since p + q cannot be greater than unity, the Statement is wrong

Let X be a random variable follows binomial distribution with probability value p = 1/2 and q = 1/2
probability that exactcy 2 heads occur are as follows
P(X = 2) = \(\left( \begin{matrix} 6 \\ x \end{matrix} \right) { p }^{ x }{ q }^{ n-x }\)
\(=\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ { \left( \frac { 1 }{ 2 } \right) } }^{ 6-2 }\)
\(=\frac { 15 }{ 64 } \)