Probability card question. (without replacement.) Suppose that from a standard deck, you draw three cards without replacement. What is the expected number of black that you will draw?
The answer to this question is 1.5, however, I'm a bit confused because shouldn't a number be a whole number. Without replacement means you just take 3 cards out of the deck at random. How will the answer be 1.5.
 A: There are the same number of red and black cards in a deck, so the expectation would be that half the drawn cards will be black.
The expectation is an average. On some trials you will get 1, others 2, still others 0 or 3. The average of natural numbers need not be an integer.
A: Let $X$ be the number of black cards you'll draw (a random variable). $X$ can take on values 0, 1, 2, or 3.
$\mathbb{P}(X=0) = 26/52 \times 25/51 \times 24/50$ (all 3 cards are red)
$\mathbb{P}(X=1) = 26/52 \times 26/51 \times 25/50 + 26/52 \times 26/51 \times 25/50 + 26/52 \times 25/51 \times 26/50$ (BRR or RBR or RRB)
By symmetry, $\mathbb{P}(X=2) = \mathbb{P}(X=1)$ and $\mathbb{P}(X=3) = \mathbb{P})(X=0)$
Now, let's use the definition of expectation:
$\mathbb{E}[X] = 0 \times \mathbb{P}(X=0) + 1 \times \mathbb{P}(X=1) + 2 \times \mathbb{P}(X=2) + 3 \times \mathbb{P}(X=3)$
You can simplify to get $\mathbb{E}(X) = 3/2$.
Could have gotten to this answer intuitively, but sometimes it helps to work through the details. Also, because it's an expected value, it can be a non-integer value, even though $X$ itself is always an integer!
