The probability, that a year selected at random will have 53 Mondays, is
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
A random variable $X$ has the following probability distribution
| $X=x$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| $P(X=x)$ | 0.15 | 0.23 | 0.10 | 0.12 | 0.20 | 0.08 | 0.07 | 0.05 |
For the event $E=\{X$ is a prime number $\}$, $F=\{X<4\}$, then $P(E \cup F)$ is
Let $\mathrm{A}, \mathrm{B}$ and C be three events, which are pairwise independent and $\bar{E}$ denote the complement of an event E . If $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$ and $\mathrm{P}(\mathrm{C})>0$, then $\mathrm{P}((\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) / C)$ is equal to
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