$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $P(X)$ $k$ $2k$ $4k$ $6k$ $8k$
The value of $P(1 < X < 4 \mid X \leq 2)$ is equal to:
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | $k$ | $2k$ | $4k$ | $6k$ | $8k$ |
The value of $P(1 < X < 4 \mid X \leq 2)$ is equal to:
- A$\frac{4}{7}$
- B$\frac{2}{3}$
- C$\frac{3}{7}$
- D$\frac{4}{5}$
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Similar Questions
Given that $E$ and $F$ are events such that $P(E)=0.6$,$P(F)=0.3$,and $P(E \cap F)=0.2$,find $P(E|F)$ and $P(F|E)$.
Easy
View SolutionIf $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then $P(B/A) = $
Easy
View SolutionIf $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B) \neq 1$,then $P\left( \frac{\overline{A}}{\overline{B}} \right) = $
Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
Then,
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
Then,
Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to
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