Help in solving Conditional Probability problem I appreciate the help I have received in the past. Please help solve this problem. It holds the key to my understanding of other similar questions.
"A box contains 7 marbles, numbered from 1 to 7 inclusive. Three marbles are randomly drawn from the box, one at a time, and without replacement. Determine the probability that the marbles picked are alternately either odd, even odd or even, odd, even numbered."
 A: There are $4$ odd-numbered marbles and $3$ even-numbered marbles. Thus, the probability of drawing an odd-numbered marble on the first draw is $\frac47$. That leaves $3$ odd- and $3$ even-numbered marbles, so the probability of drawing an even-numbered marble at that point is $\frac36=\frac12$. That leaves $3$ odd- and $2$ even-numbered marbles, so the probability of drawing an odd-numbered marble at that point is $\frac35$. The overall probability of this sequence of events is therefore
$$\frac47\cdot\frac12\cdot\frac35=\frac6{35}\;.$$
Now reason similarly to calculate the probability of drawing an even-odd-even sequence. How should you combine these two probabilities to get the probability that you want?
A: This can also be looked at as a combinatorics problem. Count the total number of combinations of draws that have alternating parity, and take the ratio to the total number of combinations overall ($7\cdot6\cdot5$).
Take the number of odd-even-odd combinations: there are $4$ odd numbers on the first draw, $3$ even numbers on the second draw, $3$ odd numbers on the third draw.
$$4\cdot3\cdot3=36$$
Now take the number of even-odd-even combinations: $3$ even on the first draw, $4$ odd on the second draw, and $2$ even on the third draw.
$$3\cdot4\cdot2=24$$
Use the rule of sums and take the ratio:
$$\frac{36+24}{7\cdot6\cdot5}$$
