Question

Here it is: You buy a six-pack of bottled Pepsi, at a time when Pepsi is running a promotion on their bottled items: Underneath every bottle cap, Pepsi has printed a message of either "Winner", or "Please Play Again". Pepsi prints "Winner" underneath 10% of all the bottle caps. a) What is the probability that you will win, meaning that at least one of the bottles will have "Winner" under its cap? b) What minimum number of individual bottles must you purchase (not necessarily in six- packs!), in order that you will have at least 97% chance of winning?

Answer 1

a)

The scenario resembles that of binomial distribution,

P(X=x) = ⁿC_x p^x (1-p)^n-x

Where p = 0.1 (Proportion of Winner caps)

n = 6

P(X=0) = ⁶C₀ 0.1⁰ (1-0.1)^6-0 = 0.531

Hence P(X>=1) = 1 – P(0) = 1 – 0.531 = 0.469 or 46.9%

b)

p = 0.1

q = 1 – 0.1 = 0.9

Margin of Error (E) = 100%-97% = 3%

Alpha = 0.05,

Z_Critical = 1.96

We know that,

E = Z_Critical * {p*q/n}^1/2

Solving for n we get,

n = 384.16 ~ 384