1
KCET 2020
MCQ (Single Correct Answer)
+1
-0

A die is thrown 10 times, the probability that an odd number will come up at least one time is

A
$$\frac{1}{1024}$$
B
$$\frac{1023}{1024}$$
C
$$\frac{11}{1024}$$
D
$$\frac{1013}{1024}$$
2
KCET 2020
MCQ (Single Correct Answer)
+1
-0

If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{2}$$ and $$P(A \cap B)=\frac{1}{6}$$, then $$P\left(A^{\prime} / B\right)$$ is

A
$$\frac{2}{3}$$
B
$$\frac{1}{3}$$
C
$$\frac{1}{2}$$
D
$$\frac{1}{12}$$
3
KCET 2020
MCQ (Single Correct Answer)
+1
-0

Events $$E_1$$ and $$E_2$$ from a partition of the sample space $$S$$. $$A$$ is any event such that $$P\left(E_1\right)=P\left(\dot{E}_2\right)=\frac{1}{2}, P\left(E_2 / A\right)=\frac{1}{2}$$ and $$P\left(A / E_2\right)=\frac{2}{3}$$, then $$P\left(E_1 / A\right)$$ is

A
$$\frac{1}{2}$$
B
$$\frac{2}{3}$$
C
1
D
$$\frac{1}{4}$$
4
KCET 2020
MCQ (Single Correct Answer)
+1
-0

The probability of solving a problem by three persons $$A, B$$ and $$C$$ independently is $$\frac{1}{2}, \frac{1}{4}$$ and $$\frac{1}{3}$$ respectively. Then the probability of the problem is solved by any two of them is

A
$$\frac{1}{12}$$
B
$$\frac{1}{4}$$
C
$$\frac{1}{24}$$
D
$$\frac{1}{8}$$
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