A soda manufacturer knows that sales of soda increase during the summer. To make sure that they get a large portion of the sales, they decide to offer a contest throughout the spring. The contest is pretty simple: when you open your bottle of soda, if it says, "You win!" under the cap, you get another bottle of their soda for free. The chance of winning a free soda is 1 in 12. Lucky Lucy bought five sodas and won a free one with four of the caps. What's the probability of that happening? Theoretically, how many sodas would you have to buy before having a greater than 50% chance of winning a free soda?

What's the probability of that happening?

0.022%

Theoretically, how many sodas would you have to buy before having a greater than 50% chance of winning a free soda?

25 sodas

Step-by-step explanation:

the probability of winning an extra soda = 1/12 = 0.0833

this variable has a binomial distribution where n = 5 and p = 0.0833

q = 1 - 0.0833 = 0.9167

now we need to calculate P (X = 4):

= n!/[X! x (n - X) !] x pˣ x qⁿ⁻ˣ

= 5!/[4! x 1!] x 0.0833⁴ x 0.9167¹

= (120/24) x 0.000048148 x 0.9167

= 0.00022 or 0.022%

what n would result in P (X = 4) = 0.5

0.5 = n!/[24 x (n! - 4) ] x 0.000048148 x 0.9167

0.5 = n!/24 x 0.000044137

n!/[24 x (n! - 4) ] = 0.5/0.000044137 = 11,328.2485

n!/[24 x (n! - 4) ] = 11,328

now using a calculator I will solve it by trial and error:

n! = 20

20! / (24 x 16!) = 4,845 ≤ 11,328, so n! must be higher

n! = 25

25! / (24 x 21!) = 12,650 ≥ 11,328, now I will try it with 24!

n! = 24

24! / (24 x 20!) = 10,626 ≤ 11,328, so the minimum n is 25