The probability of occurrence of an event $A$ is $1/2$ and that of $B$ is $3/10$. If $A$ and $B$ are mutually exclusive, then the probability of occurrence of neither $A$ nor $B$ is (in $/5$)
- A$4$
- B$3$
- C$2$
- D$1$
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Similar Questions
$A$ random variable $X$ has the probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| $P(X)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is
The probability of an event happening is the square of the probability of a second event,but the odds against the first are the cube of the odds against the second. The probabilities of the events are:
Medium
View SolutionGiven that the events $A$ and $B$ are such that $P(A) = \frac{1}{2}$,$P(A \cup B) = \frac{3}{5}$,and $P(B) = p$. Find $p$ if the events $A$ and $B$ are independent.
Easy
View SolutionIf $A$ and $B$ are two events such that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$ and $P(\bar{B}) = \frac{1}{3}$,then $P(A) = $
Easy
View SolutionFill in the blanks in the following table:
$P(A)$ $P(B)$ $P(A \cap B)$ $P(A \cup B)$ $0.5$ $0.35$ $.........$ $0.7$
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P(A \cup B)$ |
| $0.5$ | $0.35$ | $.........$ | $0.7$ |
Easy
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