Define the events in order to use Bayes Theorem

probability

The exercise I'm trying to solve using Bayes Theorem says: suppose you have a urn 1 with two balls: one white, one black, and a urn 2 with two black balls and two white balls. You pick randomly two balls, one at a time, from urn 2 and transfer them to urn 1. Then you pick randomly a ball from urn 1. What is the probability that the chosen ball from urn 1 is white? What is the probability that at least one white ball had been transferred?

My problem here is, how to define the events so I can use Bayes Theorem.

A = "the ball picked from urn 2 is white"
$$
P(A)=\frac{2}{4}=\frac{1}{2}
$$

I think that is the answer of the second question (the probability that at least one white ball had been transferred).

B = "the ball picked from urn 1 is white"
$$
P(B)=\frac{1}{2}
$$

Then $P(B|A)$ would be the probability of picking a white ball from urn 1 after picking a white ball in urn 2. Hence:

$$
P(B|A)=\frac{P(B\cap A)}{P(B)}=\frac{P(B)P(A)}{P(B)}=\frac{(\frac{1}{2})(\frac{1}{2})}{\frac{1}{2}}=\frac{1}{2}
$$

But I'm not sure that's the correct result… anyway, from here I understand that the Bayes Theorem must use a partition of events, and now I'm not sure what partitions should I use. Any thought will be appreciated.

Best Answer

In my opinion, the events should be

  • $X$: The number of white balls drawn from Urn 2; the three possible outcomes 2,1, and 0 define the partition you need.
  • $B$: The ball drawn from Urn 1 is white; this is the conditioning event.

The probabilities we are after are then $P(B)$ and $P(X=1|B)+ P(X=2|B)$.

Regarding the former, we use the law of total probability to say that $$P(B) = P(B|X=2)P(X=2) + P(B|X=1) P(X=1) +P(B|X=0)P(X=0).$$

Where

$$P(X=2) = 1/4$$ $$P(X=1) = 1/2$$ $$P(X=0) = 1/4$$ and where

$$P(B|X=n) = \frac{1+n}{4}$$ (If I have transferred $n$ white balls, I have $n+1$ white balls in a total of $4$.)

Once you have these, you can use Bayes' theorem to find $P(X=n|B)$ and the result:

$$P(X=1\vee X=2|B) = 1-P(X=0|B) = 1- \frac{P(B|X=0)P(X=0) }{P(B)};$$

where we have used the fact that $X=n$ is a partition: the probabilities of the events $X=1\vee X=2$ and $X=0$ sum to one $$P(X=1\vee X=2|B) + P(X=0|B) =1.$$

Note that you don't need to observe this and you can directly compute: $$P(X=1\vee X=2|B) = \frac{P(B|X=1)P(X=1) + P(B|X=2)P(X=2) }{P(B)};$$ where we have used the fact that $P(X=1\wedge X=2) =0$. The result will be the same.

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