# Probability of picking one black and one white marble

• B

TL;DR Summary
In a bowl you got 3 white and 4 black marbles.
Pick 2 at random without putting back the marble you picked.
What's the probability of you picking one white and one black marble?
My approach is the amount of successfull options / total amount of options.

I can first pick white in 3 different ways. Then black in 4 different ways
3 * 4
But I can also pick black first then white
4 * 3

Total amount of ways to pick marbles are
7 *6

So the probability is:
(3*4 + 4 * 3) / (7 * 6) = 4/7
which is correct.

But my question is, how do I know how many ways I could've picked successfull options?
In this case it¨'s obvious that I can pick 3 * 4 and 4 * 3 only, but if I have more colored marbles to pick from etc. What's the formula for that?
I don't see how it could be combinations, 7 Choose 2? or 4 Choose 1? Maybe it's binominal 2 over 1 but what does that even mean?

What sort of examples are you thinking about?

Lets say I add 5 red marbles to the bowl. What's probability of picking 3 and get one from each color?
3 * 4 * 5 / (12 * 11 * 10) * x
How would I get x? Maybe 'x = 3!' ?

Lets say I add 5 red marbles to the bowl. What's probability of picking 3 and get one from each color?
3 * 4 * 5 / (12 * 11 * 10) * x
How would I get x? Maybe 'x = 3!' ?

There are ##3!## different ways of getting one of each: BRW, BWR, RBW, RWB, WBR, WRB.

You can calculate the probability of each of these. Either by your counting method or by direct probabilities. E.g.
$$P(BRW) = \frac 4 {12} \times \frac 5 {11} \times \frac 3 {10}$$
However, if you cakculate some of the others you might notice something about all these probabilities.

If you slelect more than three marbles and are looking for, say, two reds a black and a white, then there are ##\binom 4 2 \times 2!## different ways: RRBW, RRWB, RBRW, RWRB etc.

But, again, if you start calculating the probability of each you might notice something.

sysprog
You have two picks, total.

To pick a black then a white you must pick one of 4 blacks out of 7, and then pick one of 3 whites out of 6, for a probability of 4/7 * 3/6, or 12/42.

The only other way is to first pick a white then a black, which means one of 3 whites out of 7 and then one of 4 blacks out of 6, for a probability of 3/7 * 4/6, another 12/42.

These two scenarios are disjoint and cover all possible ways of picking one of each color in two picks. So the total probability is the sum 12/42 + 12/42 = 24/42.