Bag B_1$B_{1}$ contains 6 white and 4 blue balls, Bag B_2$B_{2}$ contains 4 white and 6 blue balls, and Bag B_3$B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag B_2$B_{2}$, is:

### Core Logic Total apples = 18 (3 rotten, 15 good). Random variable X = \0, 1, 2\$X = \{0, 1, 2\}$ representing the number of rotten apples. ### Step 1: Probability Distribution P(X = 0) = frac^15C_2^18C_2 = frac105153$$P(X = 0) = \frac{^{15}C_2}{^{18}C_2} = \frac{105}{153}$$ P(X = 1) = frac^3C_1 times ^15C_1^18C_2 = frac45153$$P(X = 1) = \frac{^3C_1 \times ^{15}C_1}{^{18}C_2} = \frac{45}{153}$$ P(X = 2) = frac^3C_2^18C_2 = frac3153$$P(X = 2) = \frac{^3C_2}{^{18}C_2} = \frac{3}{153}$$ ### Step 2: Expectation E(X) = 0 times frac105153 + 1 times frac45153 + 2 times frac3153 = frac51153 = frac13$$E(X) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 2 \times \frac{3}{153} = \frac{51}{153} = \frac{1}{3}$$ ### Step 3: Variance E(X^2) = 0 times frac105153 + 1 times frac45153 + 4 times frac3153 = frac57153$$E(X^2) = 0 \times \frac{105}{153} + 1 \times \frac{45}{153} + 4 \times \frac{3}{153} = \frac{57}{153}$$ Var(X) = E(X^2) - (E(X))^2 = frac57153 - left(frac13right)^2 = frac57153 - frac17153 = frac40153$$Var(X) = E(X^2) - (E(X))^2 = \frac{57}{153} - \left(\frac{1}{3}\right)^2 = \frac{57}{153} - \frac{17}{153} = \frac{40}{153}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Probability