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Old 02-28-2013, 01:40 PM   #93
noventa
I STILL don't get it
 
Join Date: Dec 2009
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3/3 at Tim hortons. Since winning a prize and not winning a prize can be modelled a a Bernoulli Random Variable (nCx)p^x(1-p)^n-x. In n=3 events, the chance of drawing 3 successes x, is equal to 3c3*(1/6)^3(5/6)^0 with a success rate p of 1/6. Therefore, the odds of winning 3 for 3 is 1/6^3 or 1/216.
Alternatively, winning the 1/6 free food prize at Tim Hortons are independent and identical events. Therefore, if A is the event that one shall win a prize, then 3xP(A) is the probability of winning 3 prizes.

The odds of winning 4 (1/6) food prizes is 1/1296.

Who has a better streak then this?
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