Calculate probability of getting 2 red balls

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Discussion Overview

The discussion revolves around calculating the probability of drawing 2 red balls from a jar containing 10 blue balls and 2 red balls when 4 balls are drawn. Participants explore various methods and approaches to solve this probability problem, including combinations and permutations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a method for calculating the probability by listing possible combinations of red and blue balls drawn, but finds it inefficient.
  • Another participant suggests treating certain denominators as equivalent fractions to simplify calculations.
  • A participant expresses confusion about the suggested approach and considers using combinations or permutations as a shortcut.
  • One participant proposes multiplying a calculated probability value by 5, while noting that listing all combinations may not be feasible for more complex problems.
  • Another participant raises the concept of conditional probability in relation to the problem.
  • A suggestion is made to number the balls to clarify the permutations and to count those that include the red balls specifically.
  • A participant outlines a method involving counting total permutations and those that include the red balls to derive the probability.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the problem, with multiple competing views and approaches being discussed.

Contextual Notes

Some participants mention the limitations of listing combinations for more complex scenarios and the potential need for clearer methods, such as using permutations or conditional probability.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, particularly those looking for different approaches to solving problems involving combinations and permutations.

beamthegreat
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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!
 
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Look how the denominators relate in each situation. So when there is a 1, treat as 10/10 or 9/9 .
 
scottdave said:
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.
 
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.
 
Do you know conditional probability?
 
beamthegreat said:
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.
 
beamthegreat said:
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.
 

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