MHT CET 2025 22nd April Evening Shift | Probability Question 10 | Mathematics

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

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

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

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

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

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

A coin is tossed until one head appears or a tail appears 4 times in succession. The probability distribution of the number of tosses is

$$ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{8} & \frac{1}{8} & \frac{1}{2} & \frac{1}{4} \\ \hline \end{array} $$

$$ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{4} & \frac{1}{2} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$

$$ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{8} & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} \\ \hline \end{array} $$

$$ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{8} \\ \hline \end{array} $$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

The p.d.f. of a continuous random variable X is $f(x)=\left\{\begin{array}{cl}\frac{x^2}{18} & , \text { if }-3 < x < 3 \\ 0 & \text { otherwise }\end{array}\right.$

Then $\mathrm{P}[|\mathrm{X}|<2]=$

$\frac{1}{27}$

$\frac{2}{13}$

$\frac{8}{27}$

$\frac{4}{27}$

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