MHT CET · Maths · Probability
Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is
- A \(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{8}{19} & \frac{1}{285} \\ \hline \end{array}\)
- B \(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \\ \hline \end{array}\)
- C \(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{1}{285} & \frac{8}{19} \\ \hline \end{array}\)
- D \(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{285} & \frac{8}{95} & \frac{8}{19} & \frac{28}{57} \\ \hline \end{array}\)
Answer & Solution
Correct Answer
(B) \(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \\ \hline \end{array}\)
Step-by-step Solution
Detailed explanation
\( \text{Total oranges} = 4 (\text{defective}) + 16 (\text{good}) = 20 \) \( \text{Oranges drawn} = 3 \)
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