Given three identical bags each containing 10 balls, whose colours are as follows: beginarray|l|l|l|l| hline & textbfRed & textbfBlue & textbfGreen \\ hline textbfBag I & 3 & 2 & 5 \\ hline textbfBag II & 4 & 3 & 3 \\ hline textbfBag III & 5 & 1 & 4 \\ hline endarray$$\begin{array}{|l|l|l|l|} \hline & \textbf{Red} & \textbf{Blue} & \textbf{Green} \\ \hline \textbf{Bag I} & 3 & 2 & 5 \\ \hline \textbf{Bag II} & 4 & 3 & 3 \\ \hline \textbf{Bag III} & 5 & 1 & 4 \\ \hline \end{array}$$ A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p$p$ and if the ball is Green, the probability that it is from bag III is q$q$, then the value of left(frac1p + frac1qright)$\left(\frac{1}{p} + \frac{1}{q}\right)$ 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