Question

underneath the cap of brand x cola bottles is a chance to win a free cola. suppose that the probability of winning is 1 out of 11 when you buy 16 colas. what is the probability of winning at least once?

Answer 1

Probability of winning a free cola : p = 1/11

Probability of not a winning a cola : q = 1-1/11 = 10/11

Number of colas bought n : =16

X : Number times won a free cola

probability of winning at least once = P(X1)

P(X1) = 1-P(X<1) = 1-P(X=0)

P(X=0) i.e Probability of losing for all the 16 bottles bought

Probability of losing when one bottle is bought = 10/11

Probability of losing for all the 16 bottles bought =P(X=0) = (10/11)¹⁶

probability of winning at least once = 1- (10/11)¹⁶

Another Method .

we use Binomial distribution to solve .

X : Number times won a free cola;

X is a binomial distribution with n= 16 and p =1/11

And the probability mass function =

probability of winning at least once = P(X1)

P(X1) = 1-P(X<1) = 1-P(X=0)

P(X1) = 1-P(X<1) = 1-P(X=0) = 1 - (10/11)¹⁶ = 0.782371

Probability of winning at least once = P(X1) = 0.782371

Probability of winning at least once = 0.782371