Question

Assume that each of your calls to a popular radio station has a probability of 0.02 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 10th call? b. What is the probability that it requires more than five calls for you to connect? c. What is the mean number of calls needed to connect?

Answer 1

Answer:

a) 0.017

b) 0.92

c) 51

Step-by-step explanation:

Hi!

Lets define the random variable:

X = number of calls before you get connected

So if you the connect in the 10th call, X=9. If q is the probability of a busy signal, then q = 1 - p = 0.98

a) As the calls are independent, the probability of 9 consecutive calls obtaining a busy signaln and th 10th connecting is $(q^9)p=0.83*0.02=0.017$. Thus, 0.017 is the probability tha the first call that connects is the 10th.

We can generalize to:

$P(X=n) =f(n) =p(1-q)^n$

b) The probability of that it requires more tha five calls to get connected is:

$P(X\geq 5) = \sum_{n=5}^{\infty} f(n) = p \sum_{n=5}^{\infty} (1-p)^n$

We need to sum the geometric series (the formula is in allmost any text about infinite series). The sum is:

$\sum_{n=5}^{\infty} (1-p) =\sum_{n=0}^{\infty} (1-p)^n -\sum_{n=0}^{4} (1-p)^n = 1/p -\sum_{n=0}^{4} (1-p)^n$

Then:

$P(X\geq 5) = 1 - 0.02 \sum_{n=0}^{4} 0.98^n = 0.92$

c) The mean E(X) is:

$E(X) = \sum_{n=0}^{\infty} nf(n) = p\sum_{n=0}^{\infty} n(1-p)^n$

This sum can also be found in many math texts. the result is:

$E(X) = \frac{1}{p(1-p)}= 51$