1
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
$\frac{51}{101}$
2
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The p.m.f. of a random variable X is given by

$$\begin{aligned} \mathrm{P}[\mathrm{X}=x] & =\frac{\binom{5}{x}}{2^5}, \text { if } x=0,1,2,3,4,5 \\ & =0, \text { otherwise } \end{aligned}$$

Then which of the following is not correct?

A
$\mathrm{P}[\mathrm{X}=0]=\mathrm{P}[\mathrm{X}=5]$
B
$\mathrm{P}[\mathrm{X} \leq 1]=\mathrm{P}[\mathrm{X} \geq 4]$
C
$\mathrm{P}[\mathrm{X} \leq 2]=\mathrm{P}[\mathrm{X} \geq 3]$
D
$\mathrm{P}[\mathrm{X} \leq 2]>P[\mathrm{X} \geq 3]$
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If three fair coins are tossed, then variance of number of heads obtained, is

A
0.25
B
3
C
0.75
D
1.5
4
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $A$ and $B$ are two independent events such that $\mathrm{P}\left(\mathrm{A}^{\prime}\right)=0.75, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.65$ and $\mathrm{P}(\mathrm{B})=\mathrm{p}$, then value of $p$ is

A
$\frac{9}{14}$
B
$\frac{7}{15}$
C
$\frac{5}{14}$
D
$\frac{8}{15}$
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