1. ## Hypergeometrical Dist

Hi,

I need some help with this 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 ?

I started with:
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 !

2. Originally Posted by WeeG
Hi,

I need some help with this 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 ?

I started with:
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 !
Your pmf and variance are correct, but you forgot to divide 20 by 15 to get $E(X)=4/3$.

Let $U$ denote the number of 20-pen coins drawn and $V$ denote the number of 50-pen coins drawn. It is obvious that $U\sim V\sim X$. So

$P(U\leq 2) = P(U=0)+P(U=1)+P(U=2)$, and

$P(V\geq 1)=1-P(V<1)=1-P(V=0)$.

I'll let you do the actual computations.