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I got the following situation:

2 urns with 2 whites and 3 blacks
2 urns with 1 white and 4 blacks
1 urn with 4 whites and 1 blacks

A ball is drawn from an urn (and this urn is selected at random). The ball is white. Now, a second ball is drawn from the same urn. What is the probability of getting a black?

ATTEMPT: The urn is selected at random, and we have no further informations about it, so it smells of deception. I think we got to consider only the general set of 10 whites and 15 blacks. So, having independent trials, we calculate $\frac{10}{25} \times \frac{15}{24}$. (Alternatively, using conditional probability formula).

Is this correct?

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These things can be tricky, and unless one has a very well developed intuition, I would advise a formal conditional probability calculation.

The probability of white on the first pick turns out to simplify to $\frac{2}{5}$. You got that a quick way. It can also be done more slowly.

Next we compute the conditional probabilities, given white on first pick, that it came from the various urns.

The conditional probabilities turn out to be $\frac{1}{5}, \frac{1}{5},\frac{1}{10},\frac{1}{10},\frac{2}{5}$. Thus the probability that the second is black is $$\frac{1}{5}\cdot \frac{3}{4}+ \frac{1}{5}\cdot \frac{3}{4} +\frac{1}{10}\cdot \frac{4}{4} + \frac{1}{10}\cdot \frac{4}{4}+\frac{2}{5}\cdot \frac{1}{4}.$$

We leave it to you to verify that the conditional probabilities are what they are, and to fill in any missing details.

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  • $\begingroup$ Thanks. Be Ai="The first white was drawn from i-th urn". Be B="The second is white" (problem asked for a black second, but I follow you, doesnt change nothing).<br> Now we calculate conditional probability for, e.g., A1: P(A1|B)=P(intersection)/P(B). We already know P(B)=2/5, so it remains the intersection, that turns out to be P(A1)xP(B|A1), and this apllying conditional probability formula the other way; so [1/5*2/5]/(2/5)=1/5, as you pointed out. $\endgroup$
    – MadHatter
    Oct 3, 2013 at 13:02
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    $\begingroup$ I have changed from white to black. Thanks. Minor shift, complement. I did "calculate" $\Pr(C|A_1)$, and so on. For example, the probability that second is black, given that we drew from the first urn, is $\frac{3}{4}$ (there are $4$ balls left, and one of the white balls is gone). So one could call it, and write it, as a Bayes' Formula calculation. $\endgroup$
    – André Nicolas
    Oct 3, 2013 at 16:30
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    $\begingroup$ White on first pick can happen in $5$ ways; (i) chose first urn and got white; (ii) chose second and got white: and so on. Probability of white on first pick is $(1/5)(2/5)+(1/5)(2/5)+(1/5)(1/5)+(1/5)(1/5)+(1/5)(4/5)$. This simplifies to $2/5$. Standard "tree" or conditional probability calculation, the first term $(1/5)(2/5)$ is probability of choosing the first urn times (conditional) probability of white given we are selecting from first urn. Similar for all the others. $\endgroup$
    – André Nicolas
    Oct 3, 2013 at 18:30
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    $\begingroup$ You are welcome. It is important to seek clarity. $\endgroup$
    – André Nicolas
    Oct 3, 2013 at 18:46
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    $\begingroup$ Let $W$ be event white on first, and $B$ the event black on second. We want $\Pr(B|W)$. This is $\frac{\Pr(W\cap B)}{\Pr(W)}$. I take it you have no trouble with $\Pr(W)$: the event $W$ is a disjoint union of $5$ simple events. The same applies to $\Pr(W\cap B)$. The event $W\cap B$ can happen in $5$ pairwise disjoint ways. One of them is drew from first, drew white, then black. This has probability $(1/5)(2/5)(3/4)$. The other $4$ are calculated similarly. Then divide by $\Pr(W)$. $\endgroup$
    – André Nicolas
    Oct 7, 2013 at 18:48

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