Answer 1

(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 (1) + P(2) + P(3)

= k +  2k + 3k 

= 6k

= `6(1/21)`

= `2/7`