Assume that each of your calls to a popular radio station has a probability of p = 0.04 of connecting, that is, of not obtaining a busy signal. Assume that your calls are independent.

(a) What is the probability that your first call that connects is your tenth call? Round your answer to four decimal places (e. g. 98.7654).

(b) What is the probability that it requires more than five calls for you to connect? Round your answer to four decimal places (e. g. 98.7654).

(c) What is the mean number of calls needed to connect? Round your answer to two decimal places (e. g. 98.76).

i have no such idea

Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}