MHT CET 2025 22nd April Evening Shift | Probability Question 1 | Mathematics

If a random variable $X$ has p.d.f. $f(x)=\left\{\begin{array}{ll}\frac{a x^2}{2}+b x & , \text { if } 1 \leqslant x \leqslant 3 \\ 0 & , \text { otherwise }\end{array}\right.$ and $f(2)=2$, then the values of $a$ and $b$ are, respectively

$11,-10$

$\quad-9,10$

$\frac{1}{6}, \frac{5}{6}$

$9,-8$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

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