Question

The probability distribution of a random variable x is given as under:
P(X = x) = `{{:("k"x^2,  "for"  x = 1"," 2"," 3),(2"k"x,  "for"  x = 4"," 5"," 6),(0,  "otherwise"):}`
where k is a constant. Calculate P(X ≥ 4)

Answer 1

Given that: P(X = x) = `{{:("k"x^2,  "for"  x = 1"," 2"," 3),(2"k"x,  "for"  x = 4"," 5"," 6),(0,  "otherwise"):}`

∴ Probability distribution of random variable X is

X	1	2	3	4	5	6	otherwise
P(X)	k	4k	9k	8k	10k	12k	0

We know that `sum_("i" = 1)^"n" "P"("X"_"i")` = 1

∴ k + 4k + 9k + 8k + 10k + 12k = 1

⇒ 44k = 1

⇒ k = `1/44`

P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= 8k + 10k + 12k = 30k

= `30 xx 1/44`

= `15/22`.