Help in solving Conditional Probability problem

Question

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."

I take it that for odd we have 1,3,5,7 and for even 2,4,6 and that by picking each of the 3 individual marbles, subsesequent sample space reduces. is that true? — Sylvester, Aug 05 '13 at 15:45
Yes, it is: $1,3,5$, and $7$ are the odd numbers in the specified range of numbers, and without replacement means that once a marble has been drawn, it is no longer available in the box to be drawn again. — Brian M. Scott, Aug 05 '13 at 15:52

score 1 · Accepted Answer · answered Aug 05 '13 at 15:51

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?

many thanks. think for E-O-E should be 3/7 . 4/6 . 2/5. I appreciate. — Sylvester, Aug 05 '13 at 16:27
@Sylvester: Yes, that’s the right product. You’re welcome! — Brian M. Scott, Aug 05 '13 at 16:34

score 0 · Answer 2 · answered Aug 05 '13 at 16:27

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:

$$\frac{36+24}{7\cdot6\cdot5}$$

Help in solving Conditional Probability problem

2 Answers2