Question

For what values of k can f(x) = (1 k)k x serve as the probability distribution of a random variable x when x = 0, 1, 2, . . .

Answer 1

Hey buddy here is ur answer !!!

1st condition ,

f(0) = ( k ) k (0)
f(0) = 0 k^2

2nd Condition ,

f(1) = k (k)1
f(1) = 1k^2

3rd condition ,

f(2) = 1k (k)2
f(2) = 2k^2


Hope u like my process !!

$BE BRAINLY$

Answer 2

Concept:

A probability distribution is a statistical function that describes all of the potential values and probabilities for a random variable within a particular range.

Given:

$f(x)=(1 k)kx\\when x= 0,1,2,...$

To Find:

The value of k

Solution:

A probability distribution is a statistical function that describes all of the potential values and probabilities for a random variable within a particular range.

We are given that x= 0. 1, 2,....

So, to find the value of k we will put the values of x one by one in f(x)

$x=0\\f(x)=k*0\\x=0\\x=1\\f(x)=k*1\\=k\\x=2\\f(x)=k*2\\=2k \\x=n\\f(x)=nk$

So, now  ∑ f(x) =

$0+ k+2k+.....nk=1\\k(1+2+....+n)=1\\k(\frac{n(n+1)}{2})=1\\ k=\frac{1}{\frac{n(n+1)}{2} } \\k=\frac{2}{n(n+1)}$

∵ (1+2+....n)=$\frac{n(n+1)}{2}$

Hence, the value of k is $\frac{2}{n(n+1)}$