MHT CET 2024 2nd May Morning Shift | Probability Question 58 | Mathematics

Ten bulbs are drawn successively, with replacement, from a lot containing $10 \%$ defective bulbs, then the probability that there is at least one defective bulb, is

$1-\left(\frac{1}{10}\right)^{10}$

$1-\left(\frac{3}{10}\right)^{10}$

$1-\left(\frac{9}{10}\right)^{10}$

$1-\left(\frac{7}{10}\right)^{10}$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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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

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

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

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

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

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