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Question
The probability distribution of discrete r.v. X is as follows :
| x = x | 1 | 2 | 3 | 4 | 5 | 6 |
| P[x=x] | k | 2k | 3k | 4k | 5k | 6k |
(i) Determine the value of k.
(ii) Find P(X≤4), P(2<X< 4), P(X≥3).
Solution
(i) Since P (x) is a probability distribution of x,
`Sigma_(x = 0)^6 P(x) = 1`
∴ P (0) + P (1) + P(2) + P(3) + P(4) + P(5) + P (6) = 1
∴ k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1
∴ k = `1/21`
(ii)
P(X ≤ 4)
= P (0) + P (1) + P(2) + P(3) + P(4)
=k + 2k + 3k + 4k
= `10k`
= `10 (1/21)`
= `10/21`
P(2< X < 4)
= P(3)
= `3k`
= `3(1/21)`
= `1/7`
P(X ≥ 3)
= P (3) + P(4) + P(5) + P(6)
= `3k + 4k + 5k + 6k`
= `18k`
= `18×1/21`= `6/7`
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