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Exercise in Probability!!!

Sun 14 Apr, 2013 04:12 am
Hey guys!!!I have you can help me at the followining exercise:
We have a container that contains r balls that have numbers 1,...,r. We choose at random n of them without replacement. Let Y be the greatest and Z the smallest of the numbers of the balls we chose. Which are the probabilities P(Y<=y) and P(Z>=z)???
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engineer

2
Sun 14 Apr, 2013 05:43 am
@evinda,
P(Y<=y) = C(y,n)/C(r,n)
P(Z>=x) = C(r-x,n)/C(r,n)
evinda

1
Sun 14 Apr, 2013 05:50 am
@engineer,
And how can I explain this???? :/
engineer

1
Sun 14 Apr, 2013 06:07 am
@evinda,
C(x,y) represents the number of combinations of x objects taken y at a time. The formula is:

C(x,y) = x!/y!/(x-y)!

If you have 20 objects, you pull two and you want to know the probability that they will be less than 15, here is the logic. You have 15 "good" balls you can pull. The number of ways to pull two balls just from those 15 is C(15,2) = 15*14/2 = 105. The total number of ways you can pull two balls from 20 balls is C(20,2) = 20*19/2 = 190. That makes the probability 105/190.
evinda

1
Mon 15 Apr, 2013 03:43 pm
@engineer,
Thank you!!!
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