### Related Formula According to Bayes' Theorem, the conditional probability of an event E_k$E_k$ given that event A$A$ has occurred is: P(E_k|A) = fracP(E_k) cdot P(A|E_k)sum_i=1^n P(E_i) cdot P(A|E_i)$P(E_k|A) = \frac{P(E_k) \cdot P(A|E_k)}{\sum_{i=1}^{n} P(E_i) \cdot P(A|E_i)}$ ### Core Logic Let A$A$ be the event that 2$2$ white and 2$2$ black balls are drawn from the bag containing 8$8$ balls. Since 2$2$ white and 2$2$ black balls have already been drawn, the initial composition of the bag could only be one of the following configurations: - E_1$E_1$: 2$2$ White, 6$6$ Black - E_2$E_2$: 3$3$ White, 5$5$ Black - E_3$E_3$: 4$4$ White, 4$4$ Black (Equal composition) - E_4$E_4$: 5$5$ White, 3$3$ Black - E_5$E_5$: 6$6$ White, 2$2$ Black Assuming these 5$5$ configurations are equally likely initially: P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = frac15$P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = \frac{1}{5}$ ### Step 1: Compute Conditional Probabilities We calculate the probability of drawing 2$2$ white and 2$2$ black balls under each hypothesis using combinations: - For E_1$E_1$ (2textW, 6textB$2\text{W}, 6\text{B}$): P(A|E_1) = frac^2C_2 cdot ^6C_2^8C_4 = frac1 cdot 1570 = frac1570$P(A|E_1) = \frac{^{2}C_{2} \cdot ^{6}C_{2}}{^{8}C_{4}} = \frac{1 \cdot 15}{70} = \frac{15}{70}$ - For E_2$E_2$ (3textW, 5textB$3\text{W}, 5\text{B}$): P(A|E_2) = frac^3C_2 cdot ^5C_2^8C_4 = frac3 cdot 1070 = frac3070$P(A|E_2) = \frac{^{3}C_{2} \cdot ^{5}C_{2}}{^{8}C_{4}} = \frac{3 \cdot 10}{70} = \frac{30}{70}$ - For E_3$E_3$ (4textW, 4textB$4\text{W}, 4\text{B}$): P(A|E_3) = frac^4C_2 cdot ^4C_2^8C_4 = frac6 cdot 670 = frac3670$P(A|E_3) = \frac{^{4}C_{2} \cdot ^{4}C_{2}}{^{8}C_{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70}$ - For E_4$E_4$ (5textW, 3textB$5\text{W}, 3\text{B}$): P(A|E_4) = frac^5C_2 cdot ^3C_2^8C_4 = frac10 cdot 370 = frac3070$P(A|E_4) = \frac{^{5}C_{2} \cdot ^{3}C_{2}}{^{8}C_{4}} = \frac{10 \cdot 3}{70} = \frac{30}{70}$ - For E_5$E_5$ (6textW, 2textB$6\text{W}, 2\text{B}$): P(A|E_5) = frac^6C_2 cdot ^2C_2^8C_4 = frac15 cdot 170 = frac1570$P(A|E_5) = \frac{^{6}C_{2} \cdot ^{2}C_{2}}{^{8}C_{4}} = \frac{15 \cdot 1}{70} = \frac{15}{70}$ ### Step 2: Apply Bayes' Theorem We want to find P(E_3|A)$P(E_3|A)$, the probability that the bag contains equal numbers of white and black balls: P(E_3|A) = fracP(E_3) cdot P(A|E_3)sum_i=1^5 P(E_i) cdot P(A|E_i)$P(E_3|A) = \frac{P(E_3) \cdot P(A|E_3)}{\sum_{i=1}^{5} P(E_i) \cdot P(A|E_i)}$ P(E_3|A) = fracfrac15 cdot frac3670frac15 left( frac1570 + frac3070 + frac3670 + frac3070 + frac1570 right)$P(E_3|A) = \frac{\frac{1}{5} \cdot \frac{36}{70}}{\frac{1}{5} \left( \frac{15}{70} + \frac{30}{70} + \frac{36}{70} + \frac{30}{70} + \frac{15}{70} \right)}$ P(E_3|A) = frac3615 + 30 + 36 + 30 + 15 = frac36126 = frac27$P(E_3|A) = \frac{36}{15 + 30 + 36 + 30 + 15} = \frac{36}{126} = \frac{2}{7}$ ### Pattern Recognition Sees: Total number of balls is known, and a sample outcome is given to find the initial state distribution. Shortcut: Notice the symmetry in the configuration possibilities (E_1$E_1$ and E_5$E_5$ have identical probabilities, as do E_2$E_2$ and E_4$E_4$). Sum the numerator terms directly without writing out the full expansion to save crucial seconds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Probability