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?

 

Thanks in advance,

GaetanoH

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?

 

Thanks in advance,

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)

To receive gift --> 1/4 --> 0,25

To NOT receive gift --> 3/4 --> 0,75

 

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

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

 

1/16

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

Edited May 17, 20169 yr by Khaleesi

Thank you all and especially Eagle!

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