homework#7.data605

Q. 1

Let’s assume \[ k^n \] represents the total number of combinations.

Therefore, \[ (k-1)^n \] will be the combinations where none of the Xi are equal to 1.

\[ P(X=1) = k^n-(k-1)^n/k^n \]

For X =2 and 3, we have these:

\[ P(X=2) = (k-2+1)^n-(k-2)^n/k^n \] \[ P(X=3) = (k-3+1)^n-(k-3)^n/k^n \]

In conclusion, we get:

\[ P(X=m) = (k-m+1)^n-(k-m)^n/k^n \]

Q. 2

#a

#we know that P(success)=0.9 and p(failure)=0.1.
1-pgeom(8,0.1)

## [1] 0.3874205

#b

#we know lambda should be 1/mu, and mu is 10.
1-pexp(8,0.1)

## [1] 0.449329

#c

#we know p=0.1 and 1-p = 0.9.
pbinom(0,8,0.1)

## [1] 0.4304672

#d

#We know,
n <- 8        
p <- 1/10 

#lambda
lambda <- n * p

# success
x <- 0

poisson_p <- ((lambda^x * ((exp(1))^(-1*lambda)))/factorial(x))
poisson_p

## [1] 0.449329

#compare
ppois(0, lambda)

## [1] 0.449329