MHTCET 22 Apr 2025 Shift 2 Paper

Concept:This is a hypergeometric distribution problem, since oranges are drawn without replacement from a mixed lot containing defective and good oranges.Explanation:Let $X$ denote the number of defective oranges drawn when 3 oranges are selected.$X$ can take the values $0, 1, 2, 3$.There are $4$ defective and $16$ good oranges, so the total number of oranges is $4 + 16 = 20$.Total number of ways to choose 3 oranges from 20 is $\binom{20}{3}$.Using the hypergeometric probability formula:$P(X = r) = \frac{\binom{4}{r}\binom{16}{3-r}}{\binom{20}{3}}$, for $r = 0, 1, 2, 3$.Now calculate each probability:$P(X=0) = \frac{\binom{16}{3}}{\binom{20}{3}} = \frac{28}{57}$$P(X=1) = \frac{\binom{16}{2}\binom{4}{1}}{\binom{20}{3}} = \frac{8}{19}$$P(X=2) = \frac{\binom{16}{1}\binom{4}{2}}{\binom{20}{3}} = \frac{8}{95}$$P(X=3) = \frac{\binom{4}{3}}{\binom{20}{3}} = \frac{1}{285}$The sum of all probabilities is $1$, confirming a valid probability distribution.Answer:The probability distribution of defective oranges is:$P(X=0) = \frac{28}{57}$, $P(X=1) = \frac{8}{19}$, $P(X=2) = \frac{8}{95}$, $P(X=3) = \frac{1}{285}$.