A bag contains $4$ balls .Two balls are drawn at random and are found to be white.

What is the probability that all the four balls in the bag are white?.

The options are $\large\frac{1}{5}\space ;\frac{2}{5} ;\frac{3}{5} ;\frac{4}{5}$

My work: It's given that two balls are white.So the rest of the balls could be

  1. all white
  2. One of them white
  3. none of them white

So among the three above possibilities only one satisfies our conditions so the probability is $\frac{1}{3}$.But it not even the options given. What's wrong with my work?

  • 2
    $\begingroup$ One thing wrong in your reasoning is assuming that the three alternatives have equal probability. (why?) there are more things wrong, acutally the problem lacks information to be answered $\endgroup$
    – leonbloy
    Mar 24, 2013 at 17:21

2 Answers 2


This question cannot be answered as it is stated. You need to make some assumptions about the probabilities of the balls in the bag. How were the 4 balls chosen? Randomly between white and non-white? Randomly between white and 10 other colors?


Your list of $3$ possibilities is correct. But there is no reason to assert that these possibilities are equally likely.

The problem is poorly posed, since we are not given any information about how the balls were put into the urn. (This is technically called the prior distribution.) From the answers supplied, it is very likely that you are expected to assume that $0$ white, $1$ white, and so on up to $4$ white are (before we chose the $2$ balls) equally likely. But that is an assumption, and not even a particularly reasonable assumption.

To solve the problem carefully, we need to set up some machinery. Let $W$ be the event they are all white, and $F$ be the event the first two balls removed are white. We want $\Pr(W|F)$.

By using the defining formula $\Pr(W|F)=\dfrac{\Pr(W\cap F)}{\Pr(F)}$ for conditional probabilities, we can calculate a formal answer. If you wish, I can later complete that approach.

But let's do it more informally. We put balls into an urn $600$ times, where $120$ times we put in all white, $120$ times we put in $1$ white and $3$ black, and so on.

How many times roughly did we draw $2$ white? If we were drawing from a $2$ white $2$ black, the probability of $2$ white is $\frac{1}{2}\cdot\frac{1}{3}$, so we got $2$ white about $120/6=20$ times. If we are drawing from a $3$ white $1$ black, the probability of $2$ white is $\frac{3}{4}\cdot\frac{2}{3}$, so we got $2$ white about $60$ times. Finally, if we drew from an all white, we got $2$ white for sure, so $120$ times.

The total number of times we got $2$ white was $200$. Of these times, $120$ came from an all white situation. So the probability is $\frac{120}{200}=\frac{3}{5}$.

  • $\begingroup$ I have posted the wording of the problem exactly as in the book. And yes, I have understood the mistake in mine. $\endgroup$
    – hrkrshnn
    Mar 24, 2013 at 17:26
  • 2
    $\begingroup$ @boywholived: I assumed that the wording was verbatim. People writing elementary probability questions are all too often careless about wording. I have written out an informal solution of the problem, on the assumption, probably correct, that it is intended that we take all prior distributions of balls equally likely. This would be false if, for example, the $4$ balls were picked at random by choosing from a big bin that had equal numbers of white and black. $\endgroup$ Mar 24, 2013 at 17:42

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