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

A man takes a step forward with probability 0.4 and backwards with probability 0.6 . The probability that at the end of eleven steps, he is one step away from the starting point is

A
$${ }^{11} \mathrm{C}_6(0.24)^6$$
B
$${ }^{11} \mathrm{C}_6(0.24)^5$$
C
$${ }^{11} \mathrm{C}_6(0.4)^6(0.6)^5$$
D
$${ }^{11} \mathrm{C}_6(0.4)^5(0.6)^6$$
2
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
$$\frac{1}{4}$$
3
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
$$\frac{4}{169}$$
4
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
$$\frac{16}{169}$$
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