Question

Most sample surveys use random digit dialing equipment to call residential telephone numbers at random. The telephone polling firm Zogby International reports that the probability that a call reaches a live person is 0.25. Calls are independent.
(a) A polling firm places 7 calls. What is the probability that none of them reaches a person?


(b) When calls are made to New York City, the probability of reaching a person is only 0.06. What is the probability that none of 7 calls made to New York City reaches a person?

Answer 1

Answer:

a) $X \sim Binom(n=7, p=0.25)$  

$P(X=0)$

We can use the probability mass function and we got

$P(X=0) = (7C0) (0.25)^0 (1-0.25)^{7-0}= 0.1335$

b) $X \sim Binom(n=7, p=0.06)$  

$P(X=0)$

We can use the probability mass function and we got

$P(X=0) = (7C0) (0.06)^0 (1-0.06)^{7-0}= 0.6485$

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Part a

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=7, p=0.25)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$

And for this case we want to find this probability:

$P(X=0)$

We can use the probability mass function and we got

$P(X=0) = (7C0) (0.25)^0 (1-0.25)^{7-0}= 0.1335$

Part b

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=7, p=0.06)$  

And for this case we want to find this probability:

$P(X=0)$

We can use the probability mass function and we got

$P(X=0) = (7C0) (0.06)^0 (1-0.06)^{7-0}= 0.6485$