Question

A random variable X has the following probability distribution:

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Determine:

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

Answer 1

i. The table gives a probability distribution and therefore `sum_(i = 1)^8 P_i` = 1

∴ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

∴ 10k² + 9k – 1 = 0

∴ 10k² + 10k – k – 1 = 0

∴ 10k(k + 1) – 1(k + 1) = 0

∴ (10k – 1)(k + 1) = 0

∴ k = `1/10` or k = –1

But k cannot be negative

∴ k = `1/10`

ii. P(X < 3)

= P(X = 0 or X = 1 or X = 2)

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

= 0 + k + 2k

= 3k

= `3/10` 

iii. P(X > 4)

= P(X = 5 or X = 6 or X = 7)

= P(X = 5) + P(X = 6) + P(X = 7)

= k² + 2k² + 7k² + k

= 10k² + k

= `10(1/10)^2 + 1/10`

= `1/10 + 1/10`

= `1/5`