Random Variables

A special deck of 50 cards has 5 colors (red, white, blue, purple, and gold); for each color, there are cards numbered 1 through 10. 15 cards are picked at random without replacement from this deck. Let X be the number of purple cards picked. Find E(X) and Var(X).

Heres what I did
The probability distribution is

P(X=0) = C(40,15) / C(50,15) picks the 15 from the 40 that are not purple.

P(X=1) = C(40,14) *C(10,1) / C(50,15) picks 14 from 40 not purple and 1 from 10 purple cards.

P(X=2) = C(40,13) *C(10,2) / C(50,15) picks 13 from 40 not purple and 2 from 10 purple cards.
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P(X=10) = C(40,5) *C(10,10) / C(50,15) picks 5 from 40 not purple and 10 of 10 purple cards.
I used a calculator to find the probability of each.


Im not sure if the method i used is correct. Please comment..
The values below are the probability.
0 0.017871341971262
1 0.103103895988050
2 0.240575757305450
3 0.297855699521034
4 0.215688609997990
5 0.094902988399115
6 0.025511556021267
7 0.004100071503418
8 0.000372733773038
9 0.000017053179158
10 0.000000292340214
Method 1
Using the formula E(X) = sum P(U)*X(U) with the values above I got E(X) to be 3.0001
Var(X) =E(x^2)- [E(X)]^2
10.7143 - 3.0001^2 = 1.71
var(x) = 1.71

Method 2: Binomial random variable
E(X) = np = (15) (1/5) = 3
Var(X) = npq = (15)(1/5)(4/5) =2.4



Im not too sure about the variance..

Quote:

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Originally Posted by jjbrian

A special deck of 50 cards has 5 colors (red, white, blue, purple, and gold); for each color, there are cards numbered 1 through 10. 15 cards are picked at random without replacement from this deck. Let X be the number of purple cards picked. Find E(X) and Var(X).

Heres what I did
The probability distribution is

P(X=0) = C(40,15) / C(50,15) picks the 15 from the 40 that are not purple.

P(X=1) = C(40,14) *C(10,1) / C(50,15) picks 14 from 40 not purple and 1 from 10 purple cards.

P(X=2) = C(40,13) *C(10,2) / C(50,15) picks 13 from 40 not purple and 2 from 10 purple cards.
.
.
.
.
.
P(X=10) = C(40,5) *C(10,10) / C(50,15) picks 5 from 40 not purple and 10 of 10 purple cards.
I used a calculator to find the probability of each.

Using the formula E(X) = sum P(U)*X(U)
I got E(X) to be 3.0
then var(X) =E(x^2)- [E(X)]^2
to get var(x) to be 1.71

Im not sure if the method i used is correct. Please comment..

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I didn't check your arithmetic, but your mass function, mean and variance are all done correctly.

Good job. (Clapping)

This is hypergeometric.
Consider this as purple vs nonpurple.
You can check your calculations for the mean and variance at......
http://en.wikipedia.org/wiki/Hyperge...c_distribution
The mean is trivial.
The mean of a hypergeometric is the same as a binomial
(which is the same experiment, but you replace the cards in that case).
The variances are slightly different, but asymptotically equivalent.
The mean is np, where n=15 and p=.2, the chance of selecting a purple.
Hence the mean is 3.

I changed some stuff above. I need help finding the variance. For any random variable problems, how can you tell what method you need to use? ( binomial,independent, bernoulli's, geometric, pascal,.........)

How did you calculate the second moment? E{X^2}

E(X^2)= 0^2*0.017871341971262+ 1^2 * 0.103103895988050
+ 2^2 0.240575757305450 + 3^2 * 0.297855699521034
+4^2 * 0.215688609997990
+ 5^2 * 0.094902988399115
+ 6^2 * 0.025511556021267
+ 7^2 * 0.004100071503418
+ 8^2 * 0.000372733773038
+ 9^2 * 0.000017053179158
+ 10^2 *0.000000292340214

I calculated all this to find E(X^2) = 10.7143

Thank you all for your help. My answers are correct.
I did this problem the hard way

N=50
n=15
m=10

This is a hypergeometric distribution with: (Thanks to matheagle for the link )

E(x) = nm / N = (10)(15) / 50 = 3

V(x) = nm (N-n) (N-m) / N^2 (N-1)
V(x) = (15)(10) (50-15)(50-10) / [ 50^2 (50-1) ]
V(x) = (15)(10)(35)(40) / (2500)(49)
=1.7142