1
$\begingroup$

I have the following probability problem. I think Approach 2 is the correct approach since we need to deal with all possibilities rather than just permutations.

Question

Which is the correct approach in this probability problem?

Problem

There are 2 white and 3 red balls. A ball is selected and placed on the left chair followed by selecting another ball and placing it on the right chair. What is the probability that the left chair gets a white ball? Assume that there are only 2 chairs and you are the observer so that the left position is relative to you.

Approach 1

Total number of outcomes = 4 since possible permutations are WR or WW or RR or RW

Favorable number of outcomes = 2 since only WW or WR are possible with white ball on left.

Therefore, probability of left ball being white = 2 / 4 = .5

Approach 2

We treat each ball as a distinct object

So, Total number of outcomes = 5 X 4 = 20 since the left chair can have a ball in 5 ways and the right chair can then have it in 4 ways.

Favorable number of outcomes = 2 X 4 = 8 since the left chair can have a white ball in 2 ways and then the right chair can have any ball in 4 ways i.e. either a white ball or a red ball.

Therefore, probability of left ball being white is 8/20 = .4

  • 0
    I cannot make sense of the sentence "2 white and 3 red balls are arranged such that a ball is placed on each of the 2 chairs placed in a row.". What does it mean that a ball is placed on each of the chairs? Clearly it can only be on one of the chairs at a time? – 2017-01-23
  • 0
    Same question: Do you arrange the balls on the chairs the following way: Select a ball, place it on the left chair; select the next ball place it on the right char, and so on. What is the probability that there will be a white ball on the left chair? – 2017-01-23
  • 0
    Sorry about this. I meant a ball is placed on each chair. For example, the left chair could have a white ball placed on it and the right chair could have a red ball placed on it. – 2017-01-23
  • 0
    Zoli, yes, that's what is meant. – 2017-01-23

1 Answers 1

1

In your first approach there is a big mistake. You take every combinations with equal probabilities: $$P(WW)=P(WR)=P(RR)=P(RW)=.25$$ while the probabilities are quite differnt: $$P(WW)=\frac{2}{5}\frac{1}{4}=.1,\\ P(WR)=\frac{2}{5}\frac{3}{4}=.3,\\ P(RR)=\frac{3}{5}\frac{2}{4}=.3,\\ P(RW)=\frac{3}{5}\frac{2}{4}=.3.$$

Therefore $P(\{WW,WR\})=.1+.3=.4$

You can also see, that the second ball picking does'nt affect the forst one - if you place the white ball on the left chair, it will be white no matter what you place on the right chair. So the probability we're looking for is $$P(W)=\frac{2}{5}=.4$$

  • 0
    Ok I get it now. Thanks for your clean explanation. Would approach 2 also be a correct way? – 2017-01-23