1
$\begingroup$

There are two bowls. The first bowl contains 3 white balls and 8 black balls. The second bowl contains 6 white balls and 5 black balls. To determine the bowl that you will chose, a fair six-sided dice will be rolled. If the resulting number is greater than 4, you will randomly select a ball from the first bowl, otherwise, you will randomly select a ball from the second bowl. What's probability that you chose from the first bowl given that a black ball was chosen.

So far I have the probability of choosing from first bowl is 2/6. Probability of choosing from second bowl is 4/6.

$\endgroup$
2
$\begingroup$

Classic Bayes theorem

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

As you said $P(1)=2/6$, $P(2)=4/6$

$P(B|1)=8/11$, $P(B|2)=5/11$

And by total probability formula:

$P(B) = P(B|1)P(1)+P(B|2)P(2)$

$\endgroup$

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.