# Thread: Expected Value (Probability) Help Pls

1. ## 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.

2. 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.

3. 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? Thanks

4. 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? Thanks
Sorry, I don't know what you mean by the formula np(x).

5. 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.

6. 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.

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

7. 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.