Expected Value Problem Using Confusing Conditional Probability I am trying this question:
We have a bag with 10 blue jacks and 20 red jacks. We pick 3 jacks from the bag at random and with replacement. We are told that at least one jack is red. 
Compute the expected values of blue jacks and red jacks, again being told that we have at least one red jack
My attempt so far:
I am trying to use Bayes' law to calculate the probabilities first, before I multiply by the number of balls to find the final expected values. However, when I am trying to calculate the conditional probability,  I do not know how to include the "at least one red jack" part into the Bayes' formula. 
$$P(\text{blue jack} \mid \text{given at least one red jack}) = \frac{P(\text{blue jack}\  \cap\ \text{at least one red jack}) }{\text{P(at least one red jack)}}$$
But wouldn't the numerator of the above fraction be $0$? I am new to probability and I would appreciate some help understanding this. Thank you.

Since we draw just three jacks, and each is either red or blue,
the condition $R \geq 1$ is equivalent to $B \neq 3$.
That is, $R \geq 1$ rules out just the possible outcome $B = 3$.
This suggests the following shortcut for calculating $E(B \mid R \geq 1)$.
The unconditional expected value of the number of blue jacks is
\begin{align}
E(B)
& = 3\cdot P(B=3) + 2\cdot P(B=2) + 1
\cdot P(B=1) + 0 \cdot P(B=0) \\
& = 3\cdot P(B=3) + 2\cdot P(B=2) + P(B=1) \\
& = 1.
\end{align}
Note that we can actually calculate $E(B)=1$ by summing the expected
number of blue jacks on the first draw plus
the expected number on the second draw plus the
expected number on the third,
that is, $\frac13 + \frac13 + \frac13 = 1.$
I wrote the whole formula out only as a reminder that that formula 
also equals $E(B)$.
The conditional expectation is
$$
E(B \mid R \geq 1)
= 3\cdot P(B=3\mid R \geq 1) + 2\cdot P(B=2\mid R \geq 1) + P(B=1\mid R \geq 1).
$$
But
$$P(B=3\mid R \geq 1) = \frac{P(B=3 \cap R \geq 1)}{P(R \geq 1)} = 0,$$
while
\begin{align}
P(B=2\mid R \geq 1) &= \frac{P(B=2 \cap R \geq 1)}{P(R \geq 1)} = 
\frac{P(B=2)}{P(R \geq 1)},\\
P(B=1\mid R \geq 1) &= \frac{P(B=1 \cap R \geq 1)}{P(R \geq 1)} = 
\frac{P(B=1)}{P(R \geq 1)},
\end{align}
so
\begin{align}
E(B \mid R \geq 1)
& =  2 \frac{P(B=2 \cap R \geq 1)}{P(R \geq 1)}
      + \frac{P(B=1 \cap R \geq 1)}{P(R \geq 1)} \tag 1\\
& = \frac{1}{P(R \geq 1)}(2 P(B=2) + P(B=1)).
\end{align}
Noticing that if we just drop the first term from the formula for
$E(B)$, we get the quantity $2 P(B=2) + P(B=1)$ in the last equation,
we can write
$$2 P(B=2) + P(B=1) = E(B) - 3 P(B=3),$$
and noticing that $P(R \geq 1) = 1 - P(B = 3)$,
we can write
$$
E(B \mid R \geq 1) = \frac{1}{1 - P(B = 3)}(E(B) - 3 P(B=3)). \tag 2
$$
The right side of Equation $(2)$ should be relatively easy to calculate.
There are a lot of equations above, possibly more than if you just wrote
out the definition of $E(B \mid R \geq 1)$ and evaluated it the "hard" way,
but most of the equations are there just to show the thought process that
came up with Equation $(2)$.
I would also write at least Equation $(1)$ as part of my reasoning,
but not most of the other equations.
This isn't much of a shortcut for this particular problem, since
it's not very much more work to just work out the terms from the
definition of $E(B \mid R \geq 1)$, but if we were drawing $9$ jacks
instead of $3$ then it would make quite a difference in the
effort to get the result. We'd just be dealing with $P(B=9)$ and
$9P(B=9)$ instead of $P(B=3)$ and $3P(B=3)$.
