Probability Problem

Unknown008 · Jul 25, 2012, 12:41 AM

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 · Jul 26, 2012, 06:22 AM

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 · Jul 26, 2012, 06:31 AM

Quote:

Originally Posted by 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?

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 · Jul 27, 2012, 09:10 AM

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 · Jul 27, 2012, 09:23 AM

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

Unknown008 · Jul 27, 2012, 12:40 PM

Quote:

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.

Quote:

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 · Aug 1, 2012, 01:55 PM

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.

ebaines · Aug 1, 2012, 02:14 PM

Quote:

Originally Posted by 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 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?"

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