1
MHT CET 2023 14th May Morning Shift
+2
-0

Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens is

A
$$\mathrm{X}=x$$ 0 1 2
$$\mathrm{P[X=}x]$$ $$\frac{144}{169}$$ $$\frac{24}{169}$$ $$\frac{1}{169}$$
B
$$\mathrm{X}=x$$ 0 1 2
$$\mathrm{P[X=}x]$$ $$\frac{1}{169}$$ $$\frac{24}{169}$$ $$\frac{144}{169}$$
C
$$\mathrm{X}=x$$ 0 1 2
$$\mathrm{P[X=}x]$$ $$\frac{24}{169}$$ $$\frac{1}{169}$$ $$\frac{144}{169}$$
D
$$\mathrm{X}=x$$ 0 1 2
$$\mathrm{P[X=}x]$$ $$\frac{1}{169}$$ $$\frac{25}{169}$$ $$\frac{143}{169}$$
2
MHT CET 2023 14th May Morning Shift
+2
-0

For an initial screening of an entrance exam, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $$\frac{4}{5}$$, then the probability, that he is unable to solve less than two problems, is

A
$$\frac{201}{5}\left(\frac{1}{5}\right)^{49}$$
B
$$\frac{316}{25}\left(\frac{4}{5}\right)^{48}$$
C
$$\frac{54}{5}\left(\frac{4}{5}\right)^{49}$$
D
$$\frac{164}{25}\left(\frac{1}{5}\right)^{48}$$
3
MHT CET 2023 14th May Morning Shift
+2
-0

$$\text { If } f(x)= \begin{cases}3\left(1-2 x^2\right) & ; 0< x < 1 \\ 0 & ; \text { otherwise }\end{cases}$$ is a probability density function of $$\mathrm{X}$$, then $$\mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}\right)$$ is

A
$$\frac{75}{243}$$
B
$$\frac{23}{96}$$
C
$$\frac{179}{864}$$
D
$$\frac{52}{243}$$
4
MHT CET 2023 14th May Morning Shift
+2
-0

Three critics review a book. For the three critics the odds in favour of the book are $$2: 5, 3: 4$$ and $$4: 3$$ respectively. The probability that the majority is in favour of the book, is given by

A
$$\frac{183}{343}$$
B
$$\frac{160}{343}$$
C
$$\frac{209}{343}$$
D
$$\frac{134}{343}$$
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