SOLUTION: If a sample of 50 customers is selected on a given day, assuming that the trials are independent of each other with one outcome, what is the probability that a. fewer than 3 cus

Question 909189: If a sample of 50 customers is selected on a given day, assuming that the trials are independent of each other with one outcome, what is the probability that
a. fewer than 3 customers would receive a free newspaper 43.2?
b. 2, 3, or 4 customers would receive a free newspaper?
c. more than 5 customers would receive a free newspaper?
I have tried this multiple times using the Binomial probability formula, and my answers to not seem correct
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
You should be able to use the binomial formula on all of these and it work. However, it seems we are missing some information. We need to know the probability that a customer receives a free newspaper. Please look at your instructions again and see if you can find this vital information. For clarity, I'm going to work it out using a variable 'p' which stands for the probability of getting a free newspaper.
a) P[X<3] = P[X<=2] = P[X=0,X=1,X=2] = P[X=0] + P[X=1] + P[X=2]
(50 choose 0)(p)^0(1-p)^50 + (50 choose 1)(p)^1(1-p)^49 + (50 choose 2)(p)^2(1-p)^48
b) P[X=2] + P[X=3] + P[X=4]
(50 choose 2)(p)^2(1-p)^48 + (50 choose 3)(p)^3(1-p)^47 + (50 choose 4)(p)^4(1-p)^46
c) P[X>5] = 1-P[X<=5] = 1-[P[X<3] + P[X=3] + P[X=4] + P[X=5]]
1 - [answer to part (a) + (50 choose 3)(p)^3(1-p)^47 + (50 choose 4)(p)^4(1-p)^46 + (50 choose 5)(p)^5(1-p)^45]
Sorry that this is so general, but without the necessary information I cannot solve it as is. Hopefully this helps. If you do find the necessary information you can email me at swincher4391@yahoo.com
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