If A and B are independent events such that P(A) > 0 and P(B) > 0, then

• $\mathrm{P}\left(\mathrm{A} \cup \mathrm{B}\right) = \mathrm{P}\left(\mathrm{A}\right) . \mathrm{P}\left(\mathrm{B}\right)$

• $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \mathrm{P}\left(\mathrm{A}\right) . \mathrm{P}\left(\mathrm{B}\right)$

• $\mathrm{P}\left(\mathrm{A}|\mathrm{B}\right) = \mathrm{P}\left(\mathrm{A}\right)$

• $\mathrm{P}\left(\mathrm{B}|\mathrm{A}\right) = \mathrm{P}\left(\mathrm{B}\right)$

If A and B are two events such that P(A) = $\frac{3}{4}$ and P(B) = $\frac{5}{8}$, then

• $\mathrm{P}\left(\mathrm{A} \cup \mathrm{B}\right) \ge \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

• $\mathrm{P}\left(\mathrm{A}\text{'} \cap \mathrm{B}\right) \le 1/4$

• $\frac{3}{8} \le \mathrm{P}\left(\mathrm{A}\hspace{0.17em}\cap \mathrm{B}\right) \le \frac{5}{8}$

• All of the above

In tossing of a coin (m + n)(m > n) times, the probability of coming consecutive heads at least m times is

• $\frac{\mathrm{n} + 2}{2^{\mathrm{m} + 1}}$

• $\frac{\mathrm{m} - \mathrm{n}}{2^{\mathrm{m} + \mathrm{n}}}$

• $\frac{\mathrm{m} + \mathrm{n}}{2^{\mathrm{m} + \mathrm{n}}}$

• $\frac{\mathrm{mn}}{2^{\mathrm{m} + \mathrm{n}}}$

Two dice are thrown together. Then the probability that the sum of numbers appearing on them is a prime number, is

• $\frac{5}{12}$

• $\frac{7}{18}$

• $\frac{13}{36}$

• $\frac{11}{36}$

If the events A and B are independent, if P(A') = $\frac{2}{3}$ and P(B') = $\frac{2}{7}$, then $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right)$ is equal to

• $\frac{4}{21}$

• $\frac{5}{21}$

• $\frac{1}{21}$

• $\frac{3}{21}$

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

• ${}^{11}\mathrm{C}_5 \times {\left(0.12\right)}^5$

• ${}^{11}\mathrm{C}_5 \times {\left(0.48\right)}^5$

• ${}^{11}\mathrm{C}_6 \times {\left(0.72\right)}^6$

• ${}^{11}\mathrm{C}_6 \times {\left(0.24\right)}^5$

The probability distribution of X is

X	0	1	2	3
P(X)	0.2	k	k	2k

Find the value of k.

• 0.4

• 0.2

• 0.1

• 0.3

If P$\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \frac{7}{10} \mathrm{and} \mathrm{P}\left(\mathrm{B}\right) = \frac{17}{20}$, where P(A/B) stands for probability, then P(A/B) is equal to

• $\frac{7}{8}$

• $\frac{{\displaystyle 17}}{{\displaystyle 20}}$

• $\frac{14}{17}$

• $\frac{1}{8}$

A box has 100 pens of which 10 are defective. The probability that out of a sample of 5 pens drawn one by one with replacement and atmost one is defective, is

• $\frac{9}{10}$

• $\frac{1}{2}{\left(\frac{9}{10}\right)}^4$

• ${\left(\frac{9}{10}\right)}^5 + \frac{1}{2}{\left(\frac{9}{10}\right)}^4$

• $\frac{1}{2}{\left(\frac{9}{10}\right)}^5$

If ${}_{\mathrm{n}}\mathrm{C}_{12} = {}_{\mathrm{n}}\mathrm{C}_8$, then n is equal to

• 12

• 20

• 26

• 6