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From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?

This is what I have done. (https://i.stack.imgur.com/CZH2x.jpg)

But the answer which is given is : a/(a+b)

I can't possibly make out how it's happening.

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    $\begingroup$ Please don't post images when you can type the matter easily, as in this case. Images can't be browsed, and links can break. $\endgroup$
    – saulspatz
    Jan 24 '19 at 5:30
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    $\begingroup$ Okay I shall take care of it the next time. $\endgroup$
    – Abhishek Ghosh
    Jan 24 '19 at 15:45
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You can consider the problem as drawing balls and there is only one ball left. That's because when there are more than 1 ball left but all are the same color, it doesn't hurt to keep drawing out balls.

Also, there is the same probability of leaving any ball last.

So the probability is $\frac a{a+b}$

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Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.

This is:

$$p = \frac{\frac{(a+b-1)!}{(a-1)!b!}}{\frac{(a+b)!}{a!b!}} = \frac{(a+b-1)!}{(a-1)!b!}\cdot \frac{a!b!}{(a+b)!} = \frac{a}{a+b}$$

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  • $\begingroup$ It might seem odd asking me this. Please can you explain me as to how did you get the expression of the numerator and denominator of the probability ?? I am having a bit difficulty .. seems sort of Combination formula to me.. but how did it come ? Please do explain me the same. Thank you $\endgroup$
    – Abhishek Ghosh
    Jan 24 '19 at 15:54
  • $\begingroup$ This is based on the counting principle. The number of ways n objects can be arranged is n!. If x number of duplicates exist, then the number of ways is n!/x!. For your problem, the number of ways to arrange a and b objects is therefore (a+b)!/(a!b!). Additionally, if a always has to be the last object in the arrangement, we repeat the expression but with a - 1 instead of a. So we end up with a probability determined by dividing the number of ways where a is last by the total number of ways of arranging a+b objects. $\endgroup$
    – Phil H
    Jan 24 '19 at 18:56
  • $\begingroup$ But I think that in the probability theory problems like this demands that a white balls and b black balls are distinct. Because sample space is a set and set doesn't have repetition $\endgroup$
    – Abhishek Ghosh
    Jan 24 '19 at 19:20
  • $\begingroup$ Are you alluding to combinations versus permutations? If so, you can consider all the ways of switching white balls with each other or black balls with each other, but these samples don't look any different. Even so, the calculation then becomes (a(a+b-1))/(a+b)! reducing to the same a/(a+b). abc's solution bypassed these more complicated calculations by using the fact that the probably of any ball being last is 1/(a+b). While that is a neat and easy way of looking at this problem, it doesn't generalize to other problems, such as say, the probability of at least 2 white balls remaining. $\endgroup$
    – Phil H
    Jan 24 '19 at 20:04
  • $\begingroup$ Now i got it. And i think it shall be a.(a+b-1)!/(a+b)! $\endgroup$
    – Abhishek Ghosh
    Jan 24 '19 at 20:10

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