If we have 8 white balls and 5 black balls and one of them is picked at random, the probability of getting a white ball is 8/13. Suppose now we put 3 white balls and 2 black balls in one closed bag and the remaining balls in another identical bag. And now the experiment is that first a bag is chosen at random and then out of it, a ball is picked. Now what is the probability of getting a white ball?

Working it out one way gives one answer and working it out another way gives another ans. If we go by definition of probability here no. of outcomes in event(white balls) doesn't change (ie 8) as well as no of sample points doesn't change (ie 13).So the probability should not change (ie 8/13). But if the Total Probability Theorem is applied (considering choice of 1st bag and choice of 2nd bag as the mutually exclusive and exhaustive events), the answer is 49/80.

The book says the second one is correct (which is agreeable, though counter-intuitive). But what then is the flaw in the first method?

(if there is any other method to obtain a solution pls share)

  • $\begingroup$ Put all white balls in in a bag an all black balls in the other bag. Choose a bag and then choose a ball from the chosen bag. What is the probability that it is a white ball? $\endgroup$
    – miracle173
    Sep 17, 2015 at 5:45

3 Answers 3


Suppose we instead had six bags: five of them get a black ball each, and the last gets the eight white balls. We first pick a bag at random, then pick a ball from the bag at random. In this case the chance of getting a black ball is 5/6. The reason is that the thirteen balls are not equally likely to get picked -- the five black balls have each as much chance as getting picked as all the white balls combined. Hence a two-step process like this alters the odds from the single bag experiment.

Or, taken to extremes, suppose I tell you that in one of my hands is the winning lottery ticket (odds of winning one in 40 million) and in the other of my hands are all the losing lottery tickets (compressed in some way to fit into my hand). Obviously you would be happy to play this game, as you have a 50-50 chance of getting the winner.

  • $\begingroup$ Thanks. Your answer gives good intuition about the problem. There is one more thing I want to know- Can a sample space be defined for this experiment? If yes then what would it be? $\endgroup$ Sep 17, 2015 at 1:40
  • 1
    $\begingroup$ The sample space will be ordered pairs. Something like {(a, white), (a, black), (b, white), (b, black)}, where a and b are the labels on the bags. $\endgroup$ Sep 17, 2015 at 2:35
  • $\begingroup$ The sample space is defined as before, but individual balls now have two values attached (colour and bag), and they don't all have the same probability weight (it depends on which bag they're in). $\endgroup$ Sep 17, 2015 at 2:35
  • $\begingroup$ Here's an argument- By definition, P(E)=n(E)/n(S) and as the ans to this problem is 49/80, shouldn't the number of sample points,n(s), be at least 80? (considering that sample space is defined as the set of all elementary events that are EQUALLY likely to happen) $\endgroup$ Sep 17, 2015 at 6:29

The reason we usually assume each ball is equally likely to be chosen from a single bag is that the situation is completely symmetric with respect to what we know about the balls: they are all in the same bag and there is no reason to expect one to be more likely than another to be in the spot within the bag that you choose a ball from.

With only some of the balls in one bag, and some in the other, the situation is no longer so symmetric. We know that some balls are in a bag with four other balls, while some of the balls are in a bag with seven other balls. It is possible that the different groupings of balls in the bags will go along with different probabilities of picking each ball. Once we find out how a bag is selected, we know how the groupings of balls relate to the probabilities.


If we go by definition of probability here no. of outcomes in event(white balls) doesn't change (ie 8) as well as no of sample points doesn't change (ie 13).

That only works when the atomic outcomes are equiprobable; that is unbiased.   When you place specific numbers of each type of ball into separate bags you may introduce a bias.

If the bags are equally likely to be chosen, but there are more balls in one bag than the other, then the individual balls are not equally likely to be chosen.   Further, if the ratio of colours in the bag are not equal, then the event of selecting a colour is not independent of the choice of bag.   We then have to use the Law of Total Probability to evaluate.

Suppose we have balls labelled $\rm A, B, C$ and place ball $\rm A$ in one bag and the other two in one other.   If the choice of bag is unbiased, and the choice of ball in a bag is too, then the probability of selecting ball $\rm A$ is $1/2$: the probability of selecting the bag times the probability of selecting the only ball in that bag.   However, the probability of selecting ball $\rm B$ is $1/4$ as is that of selecting ball $\rm C$.   There's a clear atomic bias in this case.   If balls $\rm A$ and $\rm B$ are red and ball $\rm C$ is blue, then the probability of selecting a red ball is $3/4$; not $2/3$ as you would suggest.


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