B Calculate probability of getting 2 red balls

[beamthegreat]

Hi, I'm struggling with a basic probability question and I need some insight into this problem. I can solve the problem, but its a really inefficient and time consuming way.

The problem: There are 10 blue balls and 2 red balls in a jar. Calculate the probability of drawing 2 red balls if 4 balls are drawn.My solution:

The probability of getting RRBB is

2/12 * 1/11 * 1 * 1

And the probability of getting RBRB is

2/12 * 10/11 * 1/10 * 1

Then find the probability BRRB, BBRR, RBBR, and sum all of them up to get the answer.
Is there a better way to solve this problem?

Thanks!

 

[scottdave]

Look how the denominators relate in each situation. So when there is a 1, treat as 10/10 or 9/9 .

 

[beamthegreat]

Look how the denominators relate in each situation. So when there is a 1, treat as 10/10 or 9/9 .

Sorry, I don't really understand what you mean. Is there some kind of pattern I'm missing? I was thinking of shortcuts I can use like the combinations/permutation formula.

 

[beamthegreat]

OK i think I get what you mean. In all cases, the value is 0.01515151515 so I just multiply that by 5. But in harder problems I don't think listing all the possible combinations is feasible.

 

[member 587159]

Do you know conditional probability?

 

[mfb]

OK i think I get what you mean. In all cases, the value is 0.01515151515 so I just multiply that by 5. But in harder problems I don't think listing all the possible combinations is feasible.

There are more cases.
How many ways are there to choose 2 (e.g. red balls) out of 4 (total number of draws)? You can find this number without listing all cases individually.

 

[scottdave]

This may help you. https://www.mathsisfun.com/data/probability.html

 

[PeroK]

Sorry, I don't really understand what you mean. Is there some kind of pattern I'm missing? I was thinking of shortcuts I can use like the combinations/permutation formula.

One idea with these problems is to imagine that the balls are numbered. E.g. suppose the red balls are numbered 1 & 2 and the blue balls are numbered 3-12.

Note that with this approach you can see that every permutation is equally likely. E.g. 6, 8, 1, 9 is just as likely as 11, 3, 4, 12 etc.

You need any permutation that includes 1 & 2.

A) You need to count how many permutations there are.
B) You need to count how many permutations include 1 & 2.

The probability you are looking for is, therefore, B/A.

There are, of course, other ways to do these problems, but numbering balls is often a good idea to clarify things.