MHT CET 2023 12th May Morning Shift | Probability Question 18 | Mathematics

The p.m.f of random variate $$\mathrm{X}$$ is $$P(X)= \begin{cases}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \ldots, \mathrm{n} \\ 0, & \text { otherwise }\end{cases}$$

Then $$\mathrm{E}(\mathrm{X})=$$

$$\frac{\mathrm{n}+1}{3}$$

$$\frac{2 \mathrm{n}+1}{3}$$

$$\frac{\mathrm{n}+2}{3}$$

$$\frac{2 n-1}{3}$$

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

An experiment succeeds twice as often as it fails. Then the probability, that in the next 6 trials there will be atleast 4 successes, is

$$\frac{1}{729}$$

$$\frac{496}{729}$$

$$\frac{233}{729}$$

$$\frac{491}{729}$$

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then the probability distribution of number of jacks is

$$\mathrm{X}=x$$	0	1	2
$$\mathrm{P(X}=x)$$	$$\frac{144}{169}$$	$$\frac{24}{169}$$	$$\frac{1}{169}$$

$$\mathrm{X}=x$$	0	1	2
$$\mathrm{P(X}=x)$$	$$\frac{1}{169}$$	$$\frac{144}{169}$$	$$\frac{24}{169}$$

$$\mathrm{X}=x$$	0	1	2
$$\mathrm{P(X}=x)$$	$$\frac{24}{169}$$	$$\frac{1}{169}$$	$$\frac{144}{169}$$

$$\mathrm{X}=x$$	0	1	2
$$\mathrm{P(X}=x)$$	$$\frac{144}{169}$$	$$\frac{1}{169}$$	$$\frac{24}{169}$$

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is

$$\frac{1}{4}$$

$$\frac{3}{5}$$

$$\frac{2}{3}$$

$$\frac{2}{5}$$

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