# Selection without replacement

This selection without replacement question comes to me about third hand, so I'm not sure of the exact level/setting. It's definitely not my homework problem. I have done a bit of searching on the site but didn't find anything completely analogous. The question is as follows:

We have a bag with colored marbles, and the marbles also have numbers on them.

For this question all we need to know are the following:

There are 40 marbles in total.

There are 14 marbles with numbers less than or equal to three.

There are 7 purple marbles.

There is one purple marble with a number less than or equal to three.

Pulling two marbles without replacement, what is the probability the first marble has a number less than or equal to 3 and the second marble is purple?

My solution is combinatorial simply computing the probability as the number of favorable outcomes divided by the number of possible outcomes.

The denominator is easy. There are $40$ possibilities for the first draw and $39$ for the second, so $40\cdot 39=1560$ possible outcomes.

To compute the number of favorable outcomes there is only way to select a purple marble as the first marble, so there are $1\cdot 6$ favorable outcomes where the first marble is purple. That leaves $13$ ways to select the first marble if it isn't purple times the $7$ purple marbles that can then be selected on the second draw for $13\cdot 7 =91$ favorable outcomes. Taken together

$$\frac{6+91}{1560}=\frac{97}{1560}$$

for the probability the first marble has a number less than or equal to 3 and the second marble is purple.

I was wondering is my solution correct?

I would also welcome suggestions for other perhaps more probabilistic approaches to solving the problem.

• I think you are correct. You are splitting into two cases (1) first marble is purple (i.e. the "special" marble that meets both criteria) and (2) first marble is not purple. This is equivalent to using conditional probabilities but for such a simple case there is really no need to use that language. another approach is simply take $14 \times 7 = 98$ and subtract $1$ for the double-counted case of picking the special marble twice, but for a simple case like this that might take longer to visualize/explain. – antkam Apr 20 '18 at 19:53

You could do it more formally (same logic thought) in term of probability like this:

First, split the probability space onto two disjoint spaces: $\\$ $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) \\= P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) + P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ not \ purple})$

Apply Bayes formula:

$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \ P(M_1\leq3 \ \cap \ M_1 \ \mathrm{ purple}) \\ = \frac{6}{39} \frac{1}{40}$

Apply Bayes formula again:

$P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple} \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = P( M_2 \ \mathrm{ purple} | M_1\leq3 \ \cap \ M_1 \ \mathrm{ not \ purple}) P(M_1\leq3 \ \cap \ M_1 \ \mathrm{not \ purple}) \\ = \frac{7}{39}\frac{13}{40}$

Finally , $P(M_1\leq3 \ \cap \ M_2 \ \mathrm{ purple}) = \frac{6}{39} \frac{1}{40} + \frac{7}{39}\frac{13}{40} = \frac{97}{1560}$

Same result as yours (after multiple edits haha)

@antkam comment is great thought if you want to solve it without formula.