1
MHT CET 2023 9th May Evening Shift
+2
-0

A problem in statistics is given to three students A, B and C. Their probabilities of solving the problem are $$\frac{1}{2}, \frac{1}{3}$$ and $$\frac{1}{4}$$ respectively. If all of them try independently, then the probability, that problem is solved, is

A
$$\frac{2}{3}$$
B
$$\frac{3}{4}$$
C
$$\frac{1}{3}$$
D
$$\frac{1}{4}$$
2
MHT CET 2023 9th May Evening Shift
+2
-0

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards, then mean of number of queens is

A
$$\frac{1}{13}$$
B
$$\frac{1}{169}$$
C
$$\frac{2}{13}$$
D
$$\frac{4}{169}$$
3
MHT CET 2023 9th May Morning Shift
+2
-0

In a Binomial distribution with $$\mathrm{n}=4$$, if $$2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$$, then the variance is

A
$$\frac{36}{169}$$
B
$$\frac{144}{169}$$
C
$$\frac{9}{169}$$
D
$$\frac{16}{169}$$
4
MHT CET 2023 9th May Morning Shift
+2
-0

$$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ are three events, one of which must and only one can happen. The odds in favor of $$\mathrm{A}$$ are $$4: 6$$, the odds against $$B$$ are $$7: 3$$. Thus, odds against $$\mathrm{C}$$ are

A
$$7: 3$$
B
$$4: 6$$
C
$$6: 4$$
D
$$3: 7$$
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