1
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
$$\frac{1}{8}$$
2
KCET 2020
MCQ (Single Correct Answer)
+1
-0

If $$A, B, C$$ are three mutually exclusive and exhaustive events of an experiment such that $$P(A)=2 P(B)=3 P(C)$$, then $$P(B)$$ is equal to

A
$$\frac{1}{11}$$
B
$$\frac{2}{11}$$
C
$$\frac{3}{11}$$
D
$$\frac{4}{11}$$
3
KCET 2019
MCQ (Single Correct Answer)
+1
-0

Two letters are chosen from the letters of the word 'EQUATIONS'. The probability that one is vowel and the other is consonant is

A
$$\frac{3}{9}$$
B
$$\frac{8}{9}$$
C
$$\frac{5}{9}$$
D
$$\frac{4}{9}$$
4
KCET 2019
MCQ (Single Correct Answer)
+1
-0

A random variable '$$X$$' has the following probability distribution

$$x$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$
$$P(x)$$ $$k-1$$ $$3k$$ $$k$$ $$3k$$ $$3k^2$$ $$k^2$$ $$k^2+k$$

Then the value of $$k$$ is

A
$$\frac{2}{7}$$
B
$$\frac{1}{5}$$
C
$$\frac{1}{10}$$
D
$$-2$$
KCET Subjects
EXAM MAP