[MeestaD]

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:



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.

[Unknown008]

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. :)

[MeestaD]

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?

[ebaines]

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.

[MeestaD]

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)

[MeestaD]

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

[Unknown008]

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.

[Tarlika Persaud]

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.

[ebaines]

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?"