1
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable $X$ has the following probability distribution

$X=x$ 1 2 3 4 5 6 7 8
$P(X=x)$ 0.15 0.23 0.10 0.12 0.20 0.08 0.07 0.05

For the event $E=\{X$ is a prime number $\}$, $F=\{X<4\}$, then $P(E \cup F)$ is

A
0.5
B
0.77
C
0.35
D
0.75
2
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{A}, \mathrm{B}$ and C be three events, which are pairwise independent and $\bar{E}$ denote the complement of an event E . If $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$ and $\mathrm{P}(\mathrm{C})>0$, then $\mathrm{P}((\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) / C)$ is equal to

A
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{B}})$
B
$\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\overline{\mathrm{B}})$
C
$\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\mathrm{B})$
D
$\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})$
3
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable x takes the values $0,1,2$, $3, \ldots$ with probability $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant, then $\mathrm{P}(\mathrm{X}=0)$ is

A
$\frac{16}{25}$
B
$\frac{7}{25}$
C
$\frac{19}{25}$
D
$\frac{18}{25}$
4
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

One hundred identical coins, each with probability p , of showing up heads are tossed once. If $0<\mathrm{p}<1$ and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of $p$ is

A
$\frac{1}{2}$
B
$\frac{49}{101}$
C
$\frac{50}{101}$
D
$\frac{51}{101}$
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