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WeeG
04-27-2010, 10:01 AM
Hi,

got a little question regarding Hypergeometric distribution.

We mix in an urn 15 coins. 5 coins of 1 pound, 5 coins of 50 pens and 5 coins of 20 pens.
we choose randomly and without replacement 4 coins. X is defined as the number of coins of 1 pound.

a. what is the probability function of X, calculate E(X) and Var(X)
b. what is the probability to have maximum 2 coins of 20 pens ?
c. what is the probability to have at least 1 coin of 50 pens ?

On 'a' I did:
P(X=k)=((5 over k)*(10over 4-k)) / (15 over 4)
E(X)=4*5=20
Var(X)=[4*5*(15-4)*(15-5)] / [225*(15-1)]

is it correct ? how to solve b and c ? a hint would be appreciated !

BGM
04-27-2010, 10:53 AM
Part a) is correct.

You can also use Y and Z to denote the number of coins of 20 pens and 50 pens
respectively. So Y and Z are also have the identical hypergeometric distribution as X

b) The question is asking Pr{Y ≤ 2}?
= Pr{Y = 0} + Pr{Y = 1} + Pr{Y = 2} = 1 - Pr{Y = 3} - Pr{Y = 4} - Pr{Y = 5}
c) Pr{Z ≥ 1} = 1 - Pr{Z = 0}

WeeG
04-27-2010, 01:03 PM
thanks !

so in b) and c) :

if I use Y, then I treat "X" and "Z" as they were one, and when I work with Z I treat "X" and "Y" as they were one, like I did on a) ?