SOLUTION: Five out of 50 people called to give money to a charity actually agrees to give money. Based on this percentage, what is the probability that out of the next 25 people called: a.

Hi, there--

YOUR PROBLEM:
Five out of 50 people called to give money to a charity actually agrees to give money. Based 
on this percentage, what is the probability that out of the next 25 people called:

a. Exactly 1 agrees to give money?
b. At least two agree to give money?
c. At most 5 agree to give money?
d. More than 3 agree to give money?

A SOLUTION:
This situation can be modeled by the binomial distribution because we are interested in the 
number of successes in a series of independent trials (yes/no experiments). 

x = random variable = the event that a person agrees to give money
n = sample size = 25 people
p = probability of success = 5/50 = 0.10
q = probability of failure = 45/50 = 0.90

The probability distribution formula:
P(x=k) = [nCk]

a) The probability that exactly one person agrees to give money:
In this case, k = 1 so we have

P(x=1) = [25C1] 
P(x=1) = [25]*{0.0797665531]
P(x=1) = 0.1994161077


b) The probability that at least two people agree to give money:

"at least two people" means two people or more than two people. No people (x=0) or one 
person (x=1) agreeing do not qualify as "at least two." 

We calculate this probability by adding the applicable probabilities, 
P(x=2) + P(x=3) + … + P(x=24) + P(x=25). 

That's a lot of calculating. Recall that the sum of all probabilities is 1.00. Therefore, we can 
simplify our calculations by calculating 

1.00 - P(x=0) - P(x=1). 

Do you see why this works? Either method you use will give the same probability.

I'll leave it to you to set up the equation and make the calculations.


c. At most 5 agree to give money:
What does "at most 5 people" mean? Sum up the applicable probabilities just as you did in (b).

d. More than 3 agree to give money?
How many can "more than three" be? Once you determine that, you need to add the 
appropriate probabilities as you did in part (b). THere is a shortcut equation for this one too.

Good luck! Feel free to email if you have further questions about this problem.

Mrs. Figgy
math.in.the.vortex@gmail.com