Been stuck on a problem with probability. Question goes as follows:

Three balls are drawn at random without replacement from a box containing 8 white, 4 black and 4 red balls. Calculate the probabilities that there will be:

i) at least one white ball
ii) two white balls and once black ball
iii) two balls of one colour and the other of a different colour
iv) one ball of each colour

I can do i) and I calculate the inverse of it. I count 16 x 15 x 14 total outcomes with 8 ways of choosing first ball, 7 for second and 6 for third. Hence:

1 - \frac{8*7*6}{16*15*14}

Having trouble thinking my way around the others though. I think I would need to use permutations as opposed to combinations to work say ii) out as W,W,B, W,B,W and W,W,B all count as possible outcomes for the event I am interested in.

I'm starting to get confused by it all. I'd appreciate any hints or advice on how to approach this problem.

Well, you can always work your way with 'traditionally'.

WWB = 8/16 * 7/15 * 4/14
WBW = 8/16 * 4/15 * 7/14
BWW = 4/16 * 8/15 * 7/14

And then adding all three. You might immediately see a pattern here ;)

Combinations-wise:
Ways of picking 2 W = 8C2
Ways of picking 1 B = 4C1
Ways of picking any 3 balls = 16C3

The probability becomes 1/5 either way. :)

Ok, I think I see your reasoning. For question iii) though, would this involve repeating the process in ii) for all possible combinations of 2 coloured balls and 1 other and then adding them together to find the union? Or am I missing a more sophisticated way of doing this?



I’m really trying to build the link between when to use permutations/combinations for a variety of probability scenarios such as this. Does anyone know of a good resource (maybe YouTube video, document, or website page) that explains this?

Ok, I think I see your reasoning. For question iii) though, would this involve repeating the process in ii) for all possible combinations of 2 coloured balls and 1 other and then adding them together to find the union?

That's correct, but it's a little easier to think of combinations of two ball of one color and 1 ball that's not that color. In other words:

2 blacks + 1 non-black
2 wites + 1 non-white
2 reds + 1 non-red

This is a little easier than worrying all possible color combinations.

Right OK, I see that now as well. The thing is, I’m getting very confused as to when to use permutations or combinations for a probability problem. To give another example, suppose I want to know all possible outcomes of picking 3 numbers from 0,1,2,…,9 where each number can only be chosen once.

Would this be written as 10 x 9 x 8 total outcomes?

In order to calculate all possible numbers formed, say where 123, 231 and 321 are different numbers, would you use the permutations method? (10 P 3)

If interested in the collection of numbers, where 123, 321 and 231 are equivalent, would you use the combinations method? (10 C 3)

Oh, I've just realised that 10 P 3 is the same as 10 x 9 x 8.

In order to calculate all possible numbers formed, say where 123, 231 and 321 are different numbers, would you use the permutations method? (10 P 3)

Right.


If interested in the collection of numbers, where 123, 321 and 231 are equivalent, would you use the combinations method? (10 C 3)

Right again :)

Note that in both cases, you are treating numbers starting with '0', such as 012, as a valid number.

I need to calculate the following:
I have 21 different objects that need to be combined in groups of 3. The problem is that the objects are repeatable. I don't know how to do this.

I need to calculate the following:
I have 21 different objects that need to be combined in groups of 3. The problem is that the objects are repeatable. I dont know how to do this.

You need to be more specific with your question - what exactly are you trying to do? What do you mean by objects being "repeatable?"