Question

Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabilities.

x 0 1 2 3 4 5
P(X=x) 0.1 0.15 0.25 0.25 0.15 0.1

Determine the following probabilities
a. Two or three bars
b. At least one bar
c. Fewer than two bars
d. More than three bars

Answer 1

a.
`P(2 \leq X \leq 3) = 0.5`

b.
`P(X \geq 1) = 0.9`

c.
`P(X < 2) = 0.25`

d.
`P(X > 3) = 0.25`

Explanation:

We are given the following distribution:

`P(X = 0) = 0.1`

`P(X = 1) = 0.15`

`P(X = 2) = 0.25`

`P(X = 3) = 0.25`

`P(X = 4) = 0.15`

`P(X = 5) = 0.1`

a. Two or three bars

`P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) = 0.25 + 0.25 = 0.5`

Thus:

`P(2 \leq X \leq 3) = 0.5`

b. At least one bar

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1 = 0.9`

Thus:

`P(X \geq 1) = 0.9`

c. Fewer than two bars

`P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.15 = 0.25`

Thus:

`P(X < 2) = 0.25`

d. More than three bars

`P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.1 = 0.25`

Thus:

`P(X > 3) = 0.25`