1
MHT CET 2023 11th May Evening Shift
+2
-0

A fair die is tossed twice in succession. If $$\mathrm{X}$$ denotes the number of fours in two tosses, then the probability distribution of $$\mathrm{X}$$ is given by

A
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{1}{36}$$ $$\frac{25}{36}$$ $$\frac{5}{18}$$
B
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{25}{36}$$ $$\frac{1}{36}$$ $$\frac{5}{18}$$
C
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{25}{36}$$ $$\frac{5}{18}$$ $$\frac{1}{36}$$
D
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{5}{18}$$ $$\frac{1}{36}$$ $$\frac{25}{36}$$
2
MHT CET 2023 11th May Evening Shift
+2
-0

If $$\mathrm{A}$$ and $$\mathrm{B}$$ are two events such that $$\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}$$, then the value of $$\mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)$$ is

A
$$\frac{5}{6}$$
B
1
C
$$\frac{1}{6}$$
D
$$\frac{11}{6}$$
3
MHT CET 2023 11th May Evening Shift
+2
-0

Let a random variable $$\mathrm{X}$$ have a Binomial distribution with mean 8 and variance 4. If $$\mathrm{P}(\mathrm{X} \leq 2)=\frac{\mathrm{K}}{2^{16}}$$, then $$\mathrm{K}$$ is

A
17
B
121
C
136
D
137
4
MHT CET 2023 11th May Morning Shift
+2
-0

From a lot of 20 baskets, which includes 6 defective baskets, a sample of 2 baskets is drawn at random one by one without replacement. The expected value of number of defective basket is

A
0.6
B
0.06
C
0.006
D
1.07
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