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: When drawing from a bag, the probability of the third ball's color depends on what was drawn first, requiring Bayes' theorem with partition based on first draw composition.
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