MHT CET 2021 23rd September Evening Shift | Probability Question 59 | Mathematics

If the function defined by $$f(x)=K(x-x^2)$$ if $$0 < x < 1=0$$, otherwise is the p.d.f. of a r.v.X, then the value of $$P\left(X<\frac{1}{2}\right)$$ is

$$ \frac{1}{2} $$

$$ \frac{1}{3} $$

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

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

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

-0

The probability distribution of the number of doublets in four throws of a pair of dice is given by

$$\mathrm{X}$$	0	1	2	3	4
$$\mathrm{P(X) :}$$	$$\frac{1}{5}$$	$$\frac{1}{5}$$	$$\frac{1}{5}$$	$$\frac{1}{5}$$	$$\frac{1}{5}$$

$$\mathrm{X}$$	0	1	2	3
$$\mathrm{P(X) :}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$

$$\mathrm{X}$$	1	2	3	4
$$\mathrm{P(X) :}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$	$$\frac{1}{4}$$

$$\mathrm{X}$$	0	1	2	3	4
$$\mathrm{P(X) :}$$	$$\frac{625}{1296}$$	$$\frac{125}{324}$$	$$\frac{25}{216}$$	$$\frac{5}{324}$$	$$\frac{1}{1296}$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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For the probability distribution given by following

$$\mathrm{x}$$	5	6	7	8	9	10	11
$$\mathrm{P(X=x)}$$	0.07	0.2	0.3	$$\mathrm{k}$$	0.07	0.04	0.02

Var(X) =

2.65

2.85

1.65

3.85

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

-0

A random variable X has the following probability distribution

$$x$$	0	1	2	3	4	5	6	7	8
$$P(X=x)$$	K	2K	3K	4K	4K	3K	2K	K	K

Then $$\mathrm{P}(3<\mathrm{x} \leq 6)=$$

$$\frac{3}{7}$$

$$\frac{4}{7}$$

$$\frac{13}{21}$$

$$\frac{8}{21}$$

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