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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.
Solution 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` |
Solution 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` |
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