# Probabilty Question!

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I'm having some problems answering a question needed for my exams later this month.

I got this question: To promote a kind of product, the company put a gift in 1/4 of the packages. Somebody buys 2.

What is the chance he has 2 gifts, not sure wether I use binomial deviation or geometric deviation or something else?

GaetanoH

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I'm having some problems answering a question needed for my exams later this month.

I got this question: To promote a kind of product, the company put a gift in 1/4 of the packages. Somebody buys 2.

What is the chance he has 2 gifts, not sure wether I use binomial deviation or geometric deviation or something else?

GaetanoH

You should use binomial deviation, i've got the exam upcoming monday myself

You should use binomial because there's two options: You either 'win' or 'lose' (receive gift or not)

He buys 2 packages, so it has 2 trials.

Because you want to know an exact value, you'll use BinomPdf (P for precision)

So you'd have to fill in --> binomPdf(2;0,25;2) = 0,0625

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(1/4) * (1/4) = 1/16 chance

1/16

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1/4 and 1/4, which makes the chance 1 in 16 to pick both of the packages^^

Edited by Khaleesi
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Thank you all and especially Eagle!

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As was said above, this would be a solved using a binomial distribution. In this case it's quite simple as its an exact value so theres no need for cumulative tables / calculations (i.e the question isn't asking whether the customer recieved atleast 3 gifts assuming he bought like 5).

But ja, X~B(2,0.25) so you can just bang this into the formula P(X=x) = nCx * q^(n-x) * p^x (think about the formula, it makes sense logically ) note q = 1-p

apa

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