MHT CET 2023 14th May Evening Shift | Probability Question 59 | Mathematics

A fair die with numbers 1 to 6 on their faces is thrown. Let $$\mathrm{X}$$ denote the number of factors of the number, on the uppermost face, then the probability distribution of $$\mathrm{X}$$ is

$$\mathrm{X=}x$$	1	2	3	4
$$\mathrm{P(x=}x)$$	$$\frac{1}{6}$$	$$\frac{1}{2}$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$

$$\mathrm{X=}x$$	1	2	3	4
$$\mathrm{P(x=}x)$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$	$$\frac{1}{2}$$

$$\mathrm{X=}x$$	1	2	3	4
$$\mathrm{P(x=}x)$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$

$$\mathrm{X=}x$$	1	2	3	4
$$\mathrm{P(x=}x)$$	$$\frac{1}{6}$$	$$\frac{1}{6}$$	$$\frac{1}{2}$$	$$\frac{1}{6}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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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)

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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

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

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

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

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

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