A trick coin has been weighted so that the probability of heads is 0.8 and the porbability of tails is 0.2.
a) if you toss the coin 100 tims, how many heads would you expect on average?
b) What is the probability of obtaining more than 95 heads in 100 tosses?
c) What is the probability of obtaining less than 95 heads in 100 tosses?
d) What is the probability of obtaining exactly 95 heads in 100 tosses?

This is to help my GF, I don't have the book in front of me and I am not getting numbers that look right.
I got,
a) 80
b) =(0.8^96)+(0.8^97)+(0.8^98)+(0.8^99)+(0.8^100)
c) =(0.8^95)
d) = 1 - (b-answer+c-answer)

Can someone show the math to solve this? The numbers I have seem way to small and I am not sure why.

5/17/2009 6:50:14 PM

It's a binomial distribution with P=0.8. Start from there.

5/17/2009 7:03:37 PM

a) expected value = p(x)*trials = .8 * 100

b) binomial with p=.8...summation for getting 100 heads, 99 heads, 98 heads, 97 heads, and 96 heads.

c) binomail with p=.8..summation for getting 1 head, 2 heads...94 heads...same thing as 1 - summation for 95 heads, 96 heads..100 heads.

d) binomal with p=.8 ... => 100 choose 95 ...end up getting .00001497689324


**and when i say binomial..it follows the form (i think off the top of my head)

for this case (100 choose 95) * .8^95 * (1-.80)^(100-95)

[Edited on May 17, 2009 at 8:53 PM. Reason : .]

5/17/2009 8:49:44 PM

a) .8 * 100 = 80

b) (100 nCr 96)(.8)^96 * (.2)^4 + (100 nCr 97)(.8)^97 * (.2)^3 + blah blah blah

c) Find the answer to d first. Then add the answer from b to the answer from d and subtract that from 1.

d) (100 nCr 95)(.8)^95 * (.2)^5

There is a faster way to do this on a TI 83+. It has something to do a binomcdf(blah, blah, blah) but I am not sure of the syntax of what to type in.

5/18/2009 9:25:07 AM

binomcdf(100,.8,95)

gives you the sum of all the probabilities up to 95. So 1-ans would give you probability for 96,97,98,99,100.

5/18/2009 11:29:56 AM

http://www.stat.tamu.edu/~west/applets/binomialdemo.html

5/20/2009 5:10:28 PM