A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is fracmathrmmmathrmn, gcd (mathrmm,mathrmn) = 1$\frac{mathrm{m}}{mathrm{n}}, \gcd (mathrm{m},\mathrm{n}) = 1$, then mathrmn^2 -mathrmm^2$\mathrm{n}^2 -\mathrm{m}^2$ is equal to :

### 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