Binomial

[JoJo2105]

Let's say you have a binomial problem. P is 20% and q is 80%. n=100. If the question asks for the probability that a success is 95% or greater, how do you set it up? I did 100 choose 0, 1, 2, 3, 4, and 5. I added the answers together and subtracted from 1. I did not get the right answer. Can someone tell me where I am going wrong?

Thanks.

[sohpmalvin]

How do you know that you have to add up until 5?

The only way that I am aware of is to try one by one, starting from x=0 until x=m, where the probability is more than 0.05. Then take the answer to be m-1.

P(X=0)= 1.2676 E-70
P(X=1)= 5.0706 E-68
P(X=2)= 1.0039 E-65
...

Repeat this until the sum of the probability is more than 0.05.

[nostalgiajesskang]

I did 100 choose 0, 1, 2, 3, 4, and 5. I added the answers together and subtracted from 1. I did not get the right answer. Can someone tell me where I am going wrong?

1. Why do you add up until 5 only?

2. Your method is correct. You just need to add more...

[MKR]

Cant you approximate to a normal distribution?
E(x) = 100*.2=20
SD (x) = (100*.2*.8)^.5=4

I get 13 successes with a 95 % probability.

[JoJo2105]

I did 100 choose 0,1,2,3,4, and 5 because 95% of 100 is 95. I did not want to do 100 choose 100, 99, 98, 97...down to 5. So I figured I could do 0,1,2,3,4, and 5 then subtract from 1...

I don't get why that doesn't work.

[MKR]

Is the question asking for the probability of 95% successes out of 100, that is, the probability of 95 successes? Or is it asking you to find the number of successes that will happen out of 100 with a 95% certainty? (ie x number of successes will happen with 95% certainty.)

[JoJo2105]

I think it was asking for the % of successes to be atleast 95%. I remember the word ATLEAST being in there.