Question

Let X be a random variable which assumes values  x₁ , x₂, x₃ , x₄ such that  2P (X = x₁) = 3P (X = x₂) = P (X = x₃) = 5P (X = x₄). Find the probability distribution of X.

Answer 1

Let  P (X = x₁) =  k 

⇒ 2P (X = x₁) = k                      ...(i)

⇒ `P (X = x_1 ) = k/2`               ...(ii)

Parallaly P (X = x₂ ) = `k/3`         ...(iii)

And P ( X = X₄ ) = `k/5`               ...(iv)

From (i) to (iv)

`sum_(i = 1)^4 P ( X = x_i) = 1`

`k + k/2 + k/3 + k/5 = 1`

⇒ `(30k + 15k+ 10k + 6k )/30 = 1`

⇒ 61k = 30

⇒ k = `30/61`

Hence probability distribution is given by

X	x1	x2	x3	x4
P(x)	`15/61`	`10/61`	`30/61`	`6/61`

Answer 2

Let P (X = x₃) = x

P (X = x₁) = `"x"/2`

P (X = x₂) = `"x"/3`

P (X = x₄) = `"x"/5` 

`sum_(i = 1)^4"P" ( "x"_i) = 1`

P(x₁) + P(x₂) + P(x₃) + P(x₄) = 1

`"x"/2 + "x"/3 + "x" + "x"/5 = 1`

x = `30/61`

`P(X = x_1) =15/61; "P" ("X" = "x"_2) = 10/61;"P"("X" = "x"_3) = 30/61;"P" ("X" = "x"_4) = 6/61`

So, the probability distribution function will be

X	1	2	3	4
`"P"("X" = "x"_1)`	`15/61`	`10/61`	`30/61`	`6/61`