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I did find the solution to my question, and it is evidently the correct solution. However, I am not really convinced with the solution. Can someone, please help me understand and clear my doubts on this please?

The solution I found: Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be

3 B and 2 R

4 B and 1 R and

5 B and 0 R balls.

Therefore, our solution expression looks like this. 5C3 * 3C2 + 5C4 * 3C1 + 5C5 * 3C0 = 46 ways .

My Doubt: 5C3 - This means, there are 10 ways to select 3 black balls from 5 black balls, correct? But, aren't all the 10 ways same? Since all the balls are black and identical? How can there be 10 ways to select 3 balls from identically 5 black balls?

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    $\begingroup$ Yes... if you insist, the different ways of picking three black balls and two red balls can't be distinguished from any other way of picking three black balls and two red balls. However, that leads to a dead end or an incorrect answer since one might incorrectly assume that it is just as likely to have picked three black and two red balls as it is to have picked five black balls when it is not. In order to use $Pr(A)=\dfrac{|A|}{|S|}$ and use counting techniques to find probabilities, we require that the sample space in which we work be such that every outcome is equally likely to occur. $\endgroup$
    – JMoravitz
    Apr 7 '20 at 23:38
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    $\begingroup$ To get around this, we can temporarily assume that each ball is uniquely labeled, even if in reality they aren't. This allows us to work in an equiprobable sample space and can now apply counting techniques to find probabilities. $\endgroup$
    – JMoravitz
    Apr 7 '20 at 23:39
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    $\begingroup$ Compare the problem of when pulling one ball uniformly at random of finding the probability of pulling a red ball out of a bag with 99999 unlabeled black balls and one unlabeled red ball versus finding the probability of pulling a red ball out of a bag with 99999 labeled black balls and one labeled red ball. The probabilities should clearly be the same and clearly the probability for the first problem is not just $\frac{1}{2}$ despite there being only two discernible outcomes. $\endgroup$
    – JMoravitz
    Apr 7 '20 at 23:41
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Well, in my point of view, the problem asks for picking up 5 balls among the 8 there are (5 black and 3 red balls). The number ${{5}\choose{3}}$ means the different ways for selecting 3 balls from 5 no matter if are black or red. When you multiply by the factor ${{3}\choose{2}}$ is when you are forcing that those 3 balls from the five to be black.

Hope this helps you .

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