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 $\frac12$?

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.