MHT CET 2025 22nd April Evening Shift | Probability Question 8 | Mathematics

Three urns respectively contain 2 white and 3 black, 3 white and 2 black and 1 white and 4 black balls. If one ball is drawn from each um, then the probability that the selection contains 1 black and 2 white balls is

$\frac{13}{125}$

$\frac{37}{125}$

$\frac{28}{125}$

$\frac{33}{125}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

In a box containing 100 apples, 10 are defective. The probability that in a sample of 6 apples, 3 are defective is

0.1548

0.1458

0.01854

0.01458

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

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

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{8}{19} & \frac{1}{285} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{19} & \frac{8}{95} & \frac{1}{285} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{28}{57} & \frac{8}{95} & \frac{1}{285} & \frac{8}{19} \\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{285} & \frac{8}{95} & \frac{8}{19} & \frac{28}{57} \\ \hline \end{array} $$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most 2 components out of 4 tested, will survive is

$\frac{31}{3^4}$

$\frac{32}{3^4}$

$\frac{33}{3^4}$

$\frac{35}{3^4}$

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