MHT CET 2023 14th May Evening Shift | Probability Question 113 | Mathematics

The p.m.f. of a random variable $$\mathrm{X}$$ is $$\mathrm{P}(x)=\left\{\begin{array}{cl}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \mathrm{n} \\ 0, & \text { otherwise }\end{array}\right.$$, then $$\mathrm{E}(\mathrm{X})$$ is

$$\frac{\mathrm{n}+1}{6}$$

$$\frac{2 \mathrm{n}+1}{6}$$

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

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

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is a negative number, is

$$\frac{496}{1001}$$

$$\frac{505}{1001}$$

$$\frac{490}{1001}$$

$$\frac{504}{1001}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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A lot of 100 bulbs contains 10 defective bulbs. Five bulbs are selected at random from the lot and are sent to retail store. Then the probability that the store will receive at most one defective bulb is

$$\frac{7}{5}\left(\frac{9}{10}\right)^4$$

$$\frac{7}{5}\left(\frac{9}{10}\right)^5$$

$$\frac{6}{5}\left(\frac{9}{10}\right)^4$$

$$\frac{6}{5}\left(\frac{9}{10}\right)^5$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

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Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens 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
$$\mathrm{P[X=}x]$$	$$\frac{1}{169}$$	$$\frac{24}{169}$$	$$\frac{144}{169}$$

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

$$\mathrm{X}=x$$	0	1	2
$$\mathrm{P[X=}x]$$	$$\frac{1}{169}$$	$$\frac{25}{169}$$	$$\frac{143}{169}$$

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