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11 December, 03:47

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?

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  1. 11 December, 04:37
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    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
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