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.
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..
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.
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