A popular coffee shop runs a contest where prize coupons are printed under the rims of its paper cups. Suppose 10% of the cups are printed with a winning coupon, and you purchase 15 cups of coffee during the promotion. a) What type of distribution best represents this problem? Explain. b) What is the probability that you will win at least 3 prizes? c) What is the expected number of prizes you will win?
I would assume a binomial distribution http://en.wikipedia.org/wiki/Binomial_distribution
thnx i have the notes but i didn't get the answer can u help me p/z
first, what is p, the probability of winning ?
1,2, 3,4,5or 6
10% of the cups are printed with a winning coupon, try again. p is ?
p is 0.10
and the probability of k wins is \[\left(\begin{matrix}n \\k\end{matrix}\right)p^k (1-p)^{n-k}\]
What is the probability that you will win at least 3 prizes? that would be Pr(3)+Pr(4) + ... + Pr(15) or because we know Pr(0)+Pr(1)+Pr(2) + ...+ Pr(15) =1 it would be easier to figure out 1- (Pr(0) + Pr(1) + Pr(2) )
thnx is this the final answer
c) What is the expected number of prizes you will win? E(X) = n*p where n is 15 and p is 0.1
1.5
i mean for b ?
for (b) you figure out Pr(0), Pr(1), Pr(2) add them up and subtract from 1
0.1(0)+ 0.1(1)+0.1(2)-1 0.3-1 -1.7 is that true
Probabilities are always between 0 and 1. I posted the formula for Pr(k) up above. See wikipedia for details. or if you need more info, Khan is excellent. See http://www.khanacademy.org/math/probability/random-variables-topic/binomial_distribution/v/binomial-distribution-1
1-0.3=0.7 thnx so much
u really help me
Here is how to do Pr(0) \[\left(\begin{matrix}15 \\ 0\end{matrix}\right)0.1^0 0.9^{15}\] type into the google search window (15 choose 0) * 0.1^0 * 0.9^15= you get 0.205891132 for Pr(0) can you do that for Pr(1)?
i got 1.3
(15 choose 1) * 0.1^0 * 0.9^15=
(15 choose 0) * 0.1^0 * 0.9^15=0.20589113209
finlly i got it thnx
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