If $\frac{1 + 4\mathrm{p}}{4}, \frac{1 - \mathrm{p}}{4}, \frac{1 - 2\mathrm{p}}{2}$ are the probabilities of three mutually exclusive events, then

• $\frac{1}{3} \le \mathrm{p} \le \frac{1}{2}$

• $\frac{1}{3} \le \mathrm{p} \le \frac{2}{3}$

• $\frac{1}{6} \le \mathrm{p} \le \frac{1}{2}$

• None of these

If A and B are two events, then $\mathrm{P}\left(\overline{\mathrm{A}} \cap \mathrm{B}\right)$ is equal to

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

• 1 - P(A) - P(B)

• $\mathrm{P}\left(\mathrm{A}\right) + \mathrm{P}\left(\mathrm{B}\right) - \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right)$

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

Two dice are thrown simultaneously. The probability of getting a pair of ACE is

• 1/36

• 1/3

• 1/6

• None of these

If two events A and B are mutually exclusive events, then P(A/B) is equal to

• 0

• 1

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

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

If A and B are two such events that $\mathrm{P}\left(\mathrm{A} \cup \mathrm{B}\right) = \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right)$, then which of the following is true?

• P(a) + P(B) = 0

• P(A) + P(B) = P(A)P(B/A)

• P(A) + P(B) = 2P(A)P(B/A)

• None of the above

Probability of getting a total of 7 or 9 in a single throw of two dice is

• $\frac{5}{18}$

• $\frac{1}{6}$

• $\frac{1}{9}$

• None of these

Let A and B be two events of an experiment $\mathrm{P}\left(\mathrm{A}\right) = \frac{1}{4}, \mathrm{P}\left(\mathrm{A} \cup \mathrm{B}\right) = \frac{1}{2}$, then the value of $\mathrm{P}\left(\frac{\mathrm{B}}{{\mathrm{A}}^{\mathrm{C}}}\right)$ is

• $\frac{2}{3}$

• $\frac{1}{3}$

• $\frac{5}{6}$

• $\frac{1}{2}$

678.

A die is rolled three times. The probability of getting a larger number than the previous number is

• $\frac{5}{216}$

• $\frac{5}{54}$

• $\frac{1}{6}$

• $\frac{5}{36}$

If A and B are mutually exclusive events, then P(A/B) is equal to

• 0

• 1

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

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

If $\mathrm{P}\left(\mathrm{A}\right) = \frac{1}{2}, \mathrm{P}\left(\mathrm{B}\right) = \frac{1}{3} \mathrm{and} \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}\right) = \frac{1}{4}$, then the value of $\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)$ wll be

• 1

• 0

• $\frac{1}{2}$

• $\frac{1}{3}$