Conditional Probability with 2 independent sample spaces I am still grappling with probability, learning the logic and progressing one step at a time.
Please help solve this one:
A box X contains 3 VIP tickets and 4 standard tickets. Another box Y contains 7 VIP tickets and 6 standard tickets. A box is chosen at random and 2 tickets drawn from it one at a time, without replacement. Find the probability of picking a VIP and a standard ticket. 
 A: With probability $\frac{1}{2}$ we pick the first box. Let us find the probability that we got one ticket of each kind given that we picked the first box,
There are $\binom{7}{2}$ equally likely ways to choose $2$ tickets from the first box. There are $\binom{3}{1}\binom{4}{1}$ ways to pick a VIP ticket and an ordinary ticket. So the probability we pick one of each is $\frac{\binom{3}{1}\binom{4}{1}}{\binom{7}{2}}$. This is $\frac{12}{21}$, which could be simplified.
We conclude that the probability we choose the first box and pick a ticket of each kind is 
$$\frac{1}{2}\cdot\frac{12}{21}.\tag{1}$$
A similar calculation, that you should do, shows that the probability we choose the second box and pick one ticket of each kind is
$$\frac{1}{2}\cdot \frac{42}{78}.\tag{2}$$
Finally, add the two probabilities (1) and (2).
Remarks: $1.$ We give another way of finding the probability of one of each, given that we are picking from the first box. We can get one of each in two ways: (i) We pick a VIP, then an ordinary or (ii) We pick an ordinary, then a VIP.
The probability of (i) is $\frac{3}{7}\cdot\frac{4}{6}$. The probability of (ii) is $\frac{4}{7}\cdot\frac{3}{6}$. Add. We get $\frac{24}{42}$. This is the same number we obtained earlier as $\frac{12}{21}$.
The way we did the calculation in this part is perhaps simpler than the way we did it in the main answer. However, the main answer approach is in the long run more useful.
$2.$ If you have been taught to analyze multistage experiments using a tree, it would be useful to draw one. Begin by drawing two branches, each with a probability $1/2$ written beside it. These branches represent the two possible choices of box. Then calculate the probability of getting one of each for each branch. 
