Shoutbox

[code]You could also use binomial probability distribution functions or binomial cumulative distribution functions.

n = # or trials
p = P(success)
r = # of successes

in form of:
binompdf(n,p,r)    or
binomcdf(n,p,r)

n = 2
p = 0.6        {decimal of 60%)

only 1 success

binompdf as only one point is wanted

r = 1

binompdf(2,0.6,1) = 0.48 = 48%

more than 1 success

binomcdf as multiple points are wanted (in this case binompdf would also work)

I will take the cumulative probability or all integers from before 1 and take this away from total probability (1 or 100%)

r=0       {the first integer preceeding 1}

1 - binomcdf(2,0.6,0) = 1 - 0.16 = 0.84 = 84%

Just another way of proving the answer but this is more simple (only takes a couple of numbers to enter and really basic maths) and will be more valuable later when dealing with more complex questions.  Hope I helped and not confused you more, if you need more explaination just PM me.

quote:

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Originally posted by rix
Logically it would be 30%?! But heey, im not a rocket scientist.

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I can see that

*Markee runs