Probability
Probability
Allen Star Batch
Grade 12
Question:
There are two bags, bag I contains 4 red and 5 white balls, while bag II contains 5 red and 4 white balls. Two balls are drawn from bag I. If they are of the same colour, another ball is drawn from bag II. If the first two drawn balls are of different colours, one ball is drawn from bag II. If the third drawn ball is red, then the probability that the first two drawn balls were white is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - 3(p + 1) = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: This is a conditional probability problem requiring Bayes' theorem. Given that the third ball drawn is red, we need P(both first two balls were white | third ball is red) = P(third red | both white) × P(both white) / P(third red). The third ball is drawn from bag II only if the first two balls are same color; if different colors, one ball drawn from bag II with different conditional probability.
Define $E_1$ (balls from bag 1 are different colors), $E_2$ (both white), $E_3$ (both red), and $A$ (third ball is red). Calculate $P(E_2) = \frac{^5C_2}{^9C_2} = \frac{10}{36} = \frac{5}{18}$, $P(E_3) = \frac{^4C_2}{^9C_2} = \frac{6}{36} = \frac{1}{6}$, $P(E_1) = 1 - \frac{5}{18} - \frac{1}{6} = \frac{8}{9}$. Find conditional probabilities and apply Bayes' theorem: $P(E_2|A) = \frac{\frac{5}{18} \times \frac{4}{7}}{\text{total}} = \frac{45}{146}$.
Correct Answer: 8