MHT CET 2025 22nd April Morning Shift | Probability Question 31 | Mathematics

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

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

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

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

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

$\frac{1}{27}$

$\frac{2}{13}$

$\frac{8}{27}$

$\frac{4}{27}$

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

In a single toss of a fair die, the odds against the event that number 4 or 5 turns up is

$2: 1$

$1: 3$

$2: 3$

$1: 1$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

Numbers are selected at random, one at a time from the two-digit numbers $00,01,02,-------, 99$ with replacement. An event E occurs only if the product of the two digits of a selected number is 24. If four numbers are selected, then probability, that the event E occurs at least 3 times, is

$\frac{24}{(25)^4}$

$\frac{4}{(25)^4}$

$\frac{97}{(25)^4}$

$\frac{96}{(25)^4}$

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