Question

The probability distribution of X is as follows:

x	0	1	2	3	4
P[X = x]	0.1	k	2k	2k	k

Find:

1. k
2. P(X > 2)
3. P(1 < X < 4)

Solution:

The table gives a probability distribution and therefore

∑P(X = x) = 1

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

0.1 + k + 2k + 2k + k = 1

6k = 1 – 0.1

6k = 0.9

k = `square`

P(X > 2) = P(X = 3) + P(X = 4)

= 2k + k 

= `square`

P(1 < X ≤ 4 = PX = `square`) + P(X = 3) + PX = 4)

`square` + 2k + k

= 0.75

Answer 1

The table gives a probability distribution and therefore

∑P(X = x) = 1

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

0.1 + k + 2k + 2k + k = 1

6k = 1 – 0.1

6k = 0.9

k = \[\boxed{0.15}\]

P(X > 2) = P(X = 3) + P(X = 4)

= 2k + k 

= \[\boxed{0.45}\]

P(1 < X ≤ 4 = PX = \[\boxed{2}\]) + P(X = 3) + PX = 4)

\[\boxed{2k}\] + 2k + k

= 0.75