Question

Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II

Answer 1

\[
\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}
\]\(B\) is the event that they get a red ball. \(A\) is the event the get a ball from bag 2. Note that in this case \(A\subset B\) so \(\Pr(A\cap B) = \Pr(A)\)

Answer 2

Also let \(C\) be the probability they get the ball from bag 1. Getting a ball from both bags is impossible so \(A\) and \(C\) are mutually exclusive thus \(\Pr(A\cap C)=0\). Also, for \(B\) to happen, either \(A\) or \(C\) must happen.
This means: \[
\Pr(B) = \Pr(A\cup C) = \Pr(A)+\Pr(C) - \Pr(A\cap C) = \Pr(A)+\Pr(C)
\]

Answer 3

Putting this all together, we get: \[

\Pr(A|B) = \Pr(A|A\cup C) = \frac{\Pr(A)}{\Pr(A)+\Pr(C)}
\]

Answer 4

Bag I contains 3 red and 4 black balls
This means \[
\Pr(C) = \frac{3}{3+4}
\]

Bag II contains 5 red and 6 black balls.
This means \[
\Pr(A) = \frac{5}{5+6}
\]