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Urn I contains $2$ white and $4$ red balls, whereas urn II contains $1$ white and $1$ red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is

  • the probability that the ball selected from urn II is white?
  • the conditional probability that the transferred ball was white given that a white ball is selected from urn II?

I got the answer $a = \frac{4}{9}$ but I need help with $b$.
If I make $P(T) = \text{transfered ball}$ is white $P(T) = \frac13$ and P(W) = selected ball from urn 2 is white.

$P(W) = \frac49$ (from part $a$) I am looking for $P(T|W)$ by Bayes's formula - I get $$P(T|W) = P(T\cap W)/P(W)$$
then I get $$P(W|T) \cdot \dfrac{P(T)}{P(W)}$$.
we know $P(T)$ and $P(W)$ and $P(W|T) = \frac29$.
so i get $\frac29\dfrac{\frac13}{\frac49} = \frac16$ but the answer key says it's $\frac12$?

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  • $\begingroup$ We need to find $\Pr(T\cap W)$, and divide by $\Pr(W)$. The probability of $T$ is $2/6$. The probability of $W$ given $T$ is $(2/3)$. So the probability of $T\cap W$ is $(2/6)(2/3)$, which is $2/9$. Now divide by $4/9$ and simplify. $\endgroup$ Oct 2 '15 at 20:16
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In Bayes's formula formula: $P(T|W) = \frac{P(T)*P(W|T)}{P(W)}$

But $P(T)*P(W|T)=P(T\cap W)$ because these events occur at the same time, that is:

1) we have chosen white ball from urn1, probability of which is: $P(T)=\frac{2}{6}$...and now we have two white balls and one red in urn 2.

2) then we have chosen the white ball from urn 2, which means: $P(W|T)=2/3$.

So, $P(T)*P(W|T)=\frac{2}{6}*\frac{2}{3}=\frac{2}{9}=P(T\cap W)$ $$P(T|W) = P(T\cap W)/P(W)$$ We know that $P(W)=\frac{4}{9}$.

So, $P(T|W)=\frac{2}{9}/\frac{4}{9}=\frac{1}{2}$.

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  • $\begingroup$ ok, but from bayes's formula isn't P(T∩W)/P(W) equal to (P(W|T)⋅P(T)) /P (W)? so shouldn't the answers be the same? $\endgroup$
    – idknuttin
    Oct 2 '15 at 20:49
  • $\begingroup$ they are equal. $\endgroup$
    – Jane
    Oct 2 '15 at 20:53
  • $\begingroup$ look at the corrections above $\endgroup$
    – Jane
    Oct 2 '15 at 21:02
  • $\begingroup$ actually we can just use the Bayes's formula without including the probability of events intersection $\endgroup$
    – Jane
    Oct 2 '15 at 21:03

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