|
Red |
Black |
Total |
Urn I |
3 |
4 |
7 |
Urn II |
5 |
6 |
11 |
Let E1 be the event of choosing the first bag, E2 be the event of choosing the second bag.
Let A be the event of drawing a red ball.
Then P(E1) = \(\frac{1}{2}\), P(E2) = \(\frac{1}{2}\)
Also P(\(\frac{A}{E_1}\)) = P(Drawing a red ball from bag I)= \(\frac{3}{7}\)
P(\(\frac{A}{E_2}\)) = P(Drawing a red ball from bag II) = \(\frac{5}{11}\)
The probability of drawing a ball from bag I, being given that it is red, is P(\(\frac{E_1}{A}\))