Question

A random variable X has the following probability distribution:

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

Find:

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

A random variable X has the following probability distribution:

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

Determine:

1. k
2. P(X < 3)
3. P(0 < X < 3)
4. P(X > 4)

A random variable X has the following probability distribution:

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

Determine:

1. k
2. P(X < 3)
3. P(X > 6)
4. P(0 < X < 3)

A random variable X has the following probability distribution:

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

Find:

1. k
2. P(X < 3)
3. P(X > 6)

Answer 1

`sum_(i=1)^3P_i` = 1

∴ 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 = `bb((1)/(10))`

P(X < 3)

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

= 3k

= `3(1/10)`

= `3/10`

P(0 < X < 3)

= P(X = 1 or X = 2)

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

= k + 2k

= 3k

= `3(1/10)`

= `(3)/(10)`

P(X > 6)

= P(X = 7)

= 7k² + k

= `7(1/10)^2 + (1)/(10)`

= `7/100 + 1/10`

= `(17)/(100)`

P(X > 4)

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

= k² + 2k² + 7k² + k

= 10k² + k

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

= `10(1/100) + 1/10`

= `10/100 + 1/10`

= `1/10 + 1/10`

= `2/10`

= `1/5`