# Finding probability

• February 7th 2011, 11:18 AM
Marko_02
Finding probability
If a cell phone company conducted a telemarketing campaign to generate new clients, and the probability of successfully gaining a new customer was 0.08, what are the probabilities that contacting 20 potential customers would result in 0, 1, 2, 3 or 4 new customers?

I'm not sure how to get started on this question. Any help would be appreciated.
• February 7th 2011, 11:25 AM
TheEmptySet
Quote:

Originally Posted by Marko_02
If a cell phone company conducted a telemarketing campaign to generate new clients, and the probability of successfully gaining a new customer was 0.08, what are the probabilities that contacting 20 potential customers would result in 0, 1, 2, 3 or 4 new customers?

I'm not sure how to get started on this question. Any help would be appreciated.

Since the events are Bernoulli trials, that would suggest that this follows a binomial probability distribution.

$\displaystyle P(X=k) = \binom{n}{k}P^{k}(1-P)^{n-k}$

For your case $n=20,P=0.08$

Can you finish from here?
• February 7th 2011, 11:31 AM
Marko_02
thats very helpful thanks just one thing I don't know if we use different symbols but what does the k represent
• February 7th 2011, 11:34 AM
TheEmptySet
Quote:

Originally Posted by Marko_02
thats very helpful thanks just one thing I don't know if we use different symbols but what does the k represent

k is the number of new costumers.

for example $\displaystyle k=0 \implies P(X=0)$ is the probability that none of the 20 people became new costumers.
• February 7th 2011, 12:04 PM
pickslides
lets say $\displaystyle p=0.08$ and $\displaystyle n=20$

And $\displaystyle P(x=k) = ^nC_k p^k(1-p)^{n-k}$

You need to find

$\displaystyle P(x=0) = ^{20}C_0 0.08^0(1-0.08)^{20-0}=\dots$

$\displaystyle P(x=1) = ^{20}C_1 0.08^1(1-0.08)^{20-1}=\dots$

and so on...
• February 7th 2011, 12:05 PM
alexmahone
Quote:

Originally Posted by Marko_02
If a cell phone company conducted a telemarketing campaign to generate new clients, and the probability of successfully gaining a new customer was 0.08, what are the probabilities that contacting 20 potential customers would result in 0, 1, 2, 3 or 4 new customers?

I need help getting started on this question. A formula or any help would be appreciated.

Probability of not gaining a customer=1-0.08=0.92

Probability that contacting 20 potential customers would result in r new customers is $^{20}C_r(0.08)^r(0.92)^{20-r}$.
• February 7th 2011, 12:08 PM
chisigma
If we indicate with p the probability of succes in one trial, the probability to have k successes in n trials is...

$\displaystyle P(k,n) = \binom{n}{k} p^{k}\ (1-p)^{n-k}$ (1)

In your case is p=.08, n=20 and k=0,1,2,3,4...