MHTCET 22 Apr 2025 Shift 2 Paper

Concept:This is a binomial distribution problem because each apple is an independent trial with a constant probability of being defective.Explanation:Step 1: Identify the probability of selecting a defective apple:$p = \frac{10}{100} = \frac{1}{10}$Step 2: Probability of a non-defective apple:$q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$Step 3: For a sample of $n = 6$ apples, the probability that exactly $x = 3$ are defective is given by the binomial formula:$P(X = 3) = \binom{6}{3} p^3 q^{6-3}$Step 4: Substitute the values of $p$ and $q$:$P(X = 3) = \binom{6}{3} \left(\frac{1}{10}\right)^3 \left(\frac{9}{10}\right)^3$Step 5: Evaluate the binomial coefficient and powers:$\binom{6}{3} = 20$, and $\left(\frac{1}{10}\right)^3 = \frac{1}{1000}$, $\left(\frac{9}{10}\right)^3 = \frac{729}{1000}$Step 6: Multiply the terms:$P = 20 \times \frac{1}{1000} \times \frac{729}{1000} = \frac{14580}{10^6} = 0.01458$Answer:The probability is $0.01458$, which matches option D.