# Expected Value (Probability) Help Pls

• October 24th 2006, 04:46 PM
Ruichan
Expected Value (Probability) Help Pls
Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

My professor just gave the definition of expected value in yesterday's lecture then he assigned 5 questions for us to hand in tomorrow. Out of the 5 questions, I have no idea how to start or do this question.
• October 24th 2006, 06:06 PM
JakeD
Quote:

Originally Posted by Ruichan
Draw 2 cards without replacement from a deck of playing cards. What is the expected number of aces?

My professor just gave the definition of expected value in yesterday's lecture then he assigned 5 questions for us to hand in tomorrow. Out of the 5 questions, I have no idea how to start or do this question.

Here is the simplest way to answer the question.

Let the random variable X1 = 1 if an ace turns up on the first card, 0 otherwise. E(X1) = 4/52 = 1/13. Likewise let X2 =1 when an ace turns up on the second card and E(X2) = 1/13. The expected number of aces is E(X1 + X2) = E(X1) + E(X2) = 1/13 + 1/13 = 2/13. The X1 and X2 are dependent random variables but when summing expected values it does not matter.
• October 24th 2006, 06:34 PM
Ruichan
Quote:

Originally Posted by JakeD
Here is the simplest way to answer the question.

Let the random variable X1 = 1 if an ace turns up on the first card, 0 otherwise. E(X1) = 4/52 = 1/13. Likewise let X2 =1 when an ace turns up on the second card and E(X2) = 1/13. The expected number of aces is E(X1 + X2) = E(X1) + E(X2) = 1/13 + 1/13 = 2/13. The X1 and X2 are dependent random variables but when summing expected values it does not matter.

Thank you very much. I was stuck at E(X1)=1/13 coz I was thinking that without replacement, E(X2)=3/51=1/17.

Just a question on the side, how does the formula np(x) work around here or is it redundant?:confused: Thanks
• October 24th 2006, 07:04 PM
JakeD
Quote:

Originally Posted by Ruichan
Thank you very much. I was stuck at E(X1)=1/13 coz I was thinking that without replacement, E(X2)=3/51=1/17.

Just a question on the side, how does the formula np(x) work around here or is it redundant?:confused: Thanks

Sorry, I don't know what you mean by the formula np(x).
• October 24th 2006, 08:07 PM
Ruichan
Quote:

Originally Posted by JakeD
Sorry, I don't know what you mean by the formula np(x).

Sorry, wasn't clear.
It's the formula
E[X] = Summation k=1 to infinity xkp(xk)

Sorry, I can't copy and paste from my Microsoft Word equation editor. Keeps turning up blank on this. The ks are all subscripts.
• October 24th 2006, 08:58 PM
JakeD
Quote:

Originally Posted by Ruichan
Sorry, wasn't clear.
It's the formula
E[X] = Summation k=1 to infinity xkp(xk)

Sorry, I can't copy and paste from my Microsoft Word equation editor. Keeps turning up blank on this. The ks are all subscripts.

That formula applies.

$E(X_1) = \sum_{X_1=0}^1 X_1 P(X_1) = 1 \times 4/52 + 0 \times 48/52 = 4/52.$
• October 24th 2006, 09:31 PM
Ruichan
Quote:

Originally Posted by JakeD
That formula applies.

$E(X_1) = \sum_{X_1=0}^1 X_1 P(X_1) = 1 \times 4/52 + 0 \times 48/52 = 4/52.$

Thank you very much. Now I get it. I was wondering how do you apply that formula. I didn't realized that $0 \times 48/52$ has to be included.