A poker hand contains five cards find the variance of the ra
A poker hand contains five cards find the variance of the random variable of
The number of spades in a poker hand
answer is0.86397 please show how to do this
Solution
A Poker hand contanins 5 cards.
Let us consider a random variable X as the number of spades in a poker hand.
And we know that P(X = n) = P ( 5 are spades and 5 - n are not spade)
So the distribution is,
P(X = n) = [ (13 C n) * (39 C 5 - n) ] / (52 C 5) (n is form 0 to 5)
There are total 13 cards of spade.
The non spade cards are 52 - 13 = 39
Total number of cards are 52
We have to make a poker hand of 5 cards.
First we find expectation.
E(X) = n * P(X = n) (n is from 0 to 5)
For x = 0
0 * P(X = 0) = 0
For x = 1
1 * P(X = 1) = 0.41142
For x = 2
2 * P(X = 2) = 0.548559
For x = 3
3 * P(X = 3) = 0.244628
for x = 4
4 * P(X = 4) = 0.042917
for x = 5,
5 * P(X = 5) = 0.002476
Add all these probabilities so that we get E(X).
E(X) = 1.25
Now we find variance ,
variance = [ n - E(X) ]2 * P(X =n)
P(X=0) = 0.221534
P(X=1) = 0.41142
P(X=2) =0.27428
P(X=3) = 0.081543
P(X=4) = 0.010729
P(X=5) = 0.000495
For n=0
[ n - E(X) ]2 = [ 0 - 1.25]2 = 1.5625
[ n - E(X) ]2 = [ 1 -1.25]2 = 0.0625
[ n - E(X) ]2 = [ 2 - 1.25]2 = 0.5625
[ n - E(X) ]2 = [ 3 - 1.25 ] 2 = 3.0625
[ n - E(X) ]2 = [ 4 - 1.25 ] 2 = 7.5625
[ n - E(X) ]2 = [ 5 - 1.25]2 = 14.0625
variance = [ n - E(X) ]2 * P(X =n)
| P(x=n) | [n - E(x) ]^2 | multiplication | 
| 0.221534 | 1.5625 | 0.346146271 | 
| 0.41142 | 0.0625 | 0.025713723 | 
| 0.27428 | 0.5625 | 0.154282338 | 
| 0.081543 | 3.0625 | 0.249724265 | 
| 0.010729 | 7.5625 | 0.081140269 | 
| 0.000495 | 14.0625 | 0.006963723 | 
| total | 0.863970588 | |


