The probability that in a year of the 22nd century chosen at random, there will be 53 Sunday, is

• $\frac{3}{28}$

• $\frac{2}{28}$

• $\frac{7}{28}$

• $\frac{5}{28}$

If P(A) = 65, P(B) = 80, then P(A $\cap$ B) lies in the interval

• [30, 80]

• [35, 75]

• [4, 70]

• [0.45, 0.65]

For any two independent events E₁ and E₂ in a space S, $\mathrm{P}\left[\left({\mathrm{E}}_1 \cup {\mathrm{E}}_2\right) \cap \left({\mathrm{E}}_1 \cap {\mathrm{E}}_2\right)\right]$ is equal to

• $\le \frac{1}{4}$

• $> \frac{1}{4}$

• $\ge \frac{1}{2}$

• $> \frac{1}{2}$

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

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

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

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

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

115.

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

Find the value of k.

• 0.4

• 0.2

• 0.1

• 0.3