1
MHT CET 2021 23rd September Evening Shift
+2
-0

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

A
$$\frac{1}{2}$$
B
$$\frac{1}{3}$$
C
$$\frac{1}{4}$$
D
$$\frac{2}{3}$$
2
MHT CET 2021 23th September Morning Shift
+2
-0

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

A
$$\mathrm{X}$$ 0 1 2 3 4
$$\mathrm{P(X) :}$$ $$\frac{1}{5}$$ $$\frac{1}{5}$$ $$\frac{1}{5}$$ $$\frac{1}{5}$$ $$\frac{1}{5}$$
B
$$\mathrm{X}$$ 0 1 2 3
$$\mathrm{P(X) :}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$
C
$$\mathrm{X}$$ 1 2 3 4
$$\mathrm{P(X) :}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$ $$\frac{1}{4}$$
D
$$\mathrm{X}$$ 0 1 2 3 4
$$\mathrm{P(X) :}$$ $$\frac{625}{1296}$$ $$\frac{125}{324}$$ $$\frac{25}{216}$$ $$\frac{5}{324}$$ $$\frac{1}{1296}$$
3
MHT CET 2021 23th September Morning Shift
+2
-0

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) =

A
2.65
B
2.85
C
1.65
D
3.85
4
MHT CET 2021 23th September Morning Shift
+2
-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 $$\mathrm{P}(3<\mathrm{x} \leq 6)=$$

A
$$\frac{3}{7}$$
B
$$\frac{4}{7}$$
C
$$\frac{13}{21}$$
D
$$\frac{8}{21}$$
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