Question

Suppose X has probability distribution

x: 0 1 2 3 4

P(X = x) 0.2 0.1 0.2 0.2 0.3

Find the following probabilities:

a. P(X < 2)

b. P(X ≤ 2 and X < 4)

c. P(X ≤ 2 and X ≥ 1)

d. P(X = 1 or X ≤ 3)

e. P(X = 2 given X ≤ 2)

Answer 1

Solution:
X has probability distribution with table

X	P(X=x)
0	0.2
1	0.1
2	0.2
3	0.2
4	0.3

Following probabilities can be calculated as
Solution(a)
P(X<2) =? can be calculated from discrete probability distribution table as follows:
P(X<2) = P(X=0) + P(X=1) = 0.2 + 0.1 = 0.3
So there is 30% probability that X is less than 2.
Solution(b)
P(X<=2 and X<4) = ? can be calculated as
P(X<=2 and X<4) = P(X=0) + P(X=1) + P(X=2) = 0.2 + 0.1 + 0.2 = 0.5
Solution(c)
Here we need to calculate
P(X<=2 and X>=1) = ? can be calculated as
P(X<=2 and X>=1) = P(X=1) + P(X=2) = 0.1 + 0.2= 0.3

Solution(d)

P(X=1 or X<=3) = ? can be caculated as

P(X=1 or X<=3) = P(X=1) + P(X<=3) - P(X=1 and P(X<=3) = P(X=1) + P(X=0) + P(X=1) + P(X=2) + P(X=3) - P(X=1) = 0.1 + 0.2 + 0.1 + 0.2 + 0.2 - 0.1 = 0.7
Solution(e)
P(X=2 Given (X<=2) = ? can be calculated as
P(X=2 Given (X<=2) = P(X=2 and X<=2)/P(X<=2) = P(X=2)/P(X<=2) = 0.2/(0.2+0.1+0.2) = 0.2/0.5 = 0.4