Re: simple question
True. You can approximate the answer as n*p (n=#of tries, p= prob. on one try) if n and p are small. Basically, this multiply counts all the times where it happens more than once.
A conceptually different way to arrive at the accurate answer is:
P= p |chance of getting it on the first try
+(1-p)*p |chance of getting it on the second try and not getting it on the first
+(1-p)^2*p |chance of getting it on the third try and not 1st or 2nd
+(1-p)^3*p |etc.
+(1-p)^4*p
=1-(1-p)^5 |as given above with a little algebra
=52.6% for p=5/36
|