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.

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.

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

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