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

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.

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.

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

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

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) = \(\left( \begin{matrix} 6 \\ x \end{matrix} \right) { p }^{ x }{ q }^{ n-x }\)
\(=\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ { \left( \frac { 1 }{ 2 } \right) } }^{ 6-2 }\)
\(=\frac { 15 }{ 64 } \)

(i) Binomial distribution is symmetrical if p = q = 0.5. It is skew symmetric if p≠q. It is positively skewed if p < 0.5 and it is negatively skewed if p > 0.5.
(ii) For binomial distribution, variance is less than mean.
Variance = npq = (np)q < np < mean.

We are given mean μ = 2.25 and standard deviation σ = 0.25.
Probability that weight of fish is less than 2 kg is P(X < 2.0)
When x = 20 \(Z=\frac { X-\mu }{ \sigma } =\frac { 2.0-2.25 }{ 0.25 } =P(Z<-1.0)=P(Z>1.0)\)
= 0.5 – 0.3413 = 0.1587
Therefore 15.87% of fishes weigh less than 2 kg.

Let the distribution of heights be normally distributed with mean mean 68.22 and standard deviation = 3.286
\(Z=\frac { X-\mu }{ \sigma } =\frac { X-69.25 }{ 3.286 } \)
When X = 74
\(Z=\frac { X-\mu }{ \sigma } =\frac { 74-69.25 }{ 3.286 } =1.4455\)
Now P(Z > 74) = P(Z > 1.44)
= 0.5 – 0.4251
= 0.0749
Expected number of children to be over 74 cm out of 1200 children
= 1200 × 0.0749 ≈ 90 children

Let X be a random variable follows poisson distribution with number of erythrocytes.
Hence, Mean λ = 8 erythrocytes/single square
P(exactly 5 erythrocytes are in one square) = P(X = 5)=\(\\ \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -8 }{ 8 }^{ 5 } }{ 5! } \)
\(=\frac { 0.000335\times 32768 }{ 120 } \)
= 0.0916
The probability that exactly 5 erythrocytes are found in one square is 0.0916. i.e there are 9.16% chances that exactly 5 erythrocytes are found in one square.

Here p = q = 1/2
(a) probability of A beating B in exactly 3 games out of 4
\(\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 4-3 }\)
= 1/4 = 25%
(b) probability of A beating B in exactly 5 games out of 8
\(\left( \begin{matrix} 8 \\ 5 \end{matrix} \right) { (\frac { 1 }{ 2 } ) }^{ 5 }\left( \frac { 1 }{ 2 } \right) ^{ 8-5 }\)
\(=\frac { 7 }{ 32 } \) = 21.875%
Clearly, the first event is more probable.

Consider a 2,000 individuals getting injection of a given serum , n = 2000
Let X be the number of individuals suffering a bad reaction.
Let p be the probability that an individual suffers a bad reaction = 0.001
and q = 1– p = 1– 0.001 = 0.999
Since n is large and p is small, Binomial Distribtuion approximated to poisson distribution
So, λ = np = 2000 × 0.001 = 2
(i) Probability out of 2000, exactly 3 will suffer a bad reaction is
\(P(X=3)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -2 }{ 2 }^{ 3 } }{ 3! } =0.1804\)
(ii) Probability out of 2000, more than 2 individuals will suffer a bad reaction
= P(X > 2)
1-[P(X\(\le\)2)]
= 1 – [P(x = 0) + P(x = 1) + P(x = 2)]
\(=1-\left[ \frac { { e }^{ -2 }{ 2 }^{ 0 } }{ 0! } +\frac { { e }^{ -2 }{ 2 }^{ 1 } }{ 1! } +\frac { { e }^{ -2 }{ 2 }^{ 2 } }{ 2! } \right] \)
\(=1-{ e }^{ 2 }\left( \frac { { 2 }^{ 0 } }{ 0! } +\frac { { 2 }^{ 1 } }{ 1! } +\frac { { 2 }^{ 2 } }{ 2! } \right) \)
= 0.323