$A$ 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:

  • A
    $\frac{4}{7}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{3}{7}$
  • D
    $\frac{4}{5}$

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Similar Questions

Suppose $A$ and $B$ are events of a random experiment such that $P(A)=\frac{1}{3}$, $P(A \cap B)=\frac{1}{5}$ and $P(A \cup B)=\frac{3}{5}$. Match the items of List-$I$ with the items of List-$II$.
List-$I$List-$II$
$A$. $P(\frac{A}{B})$$(i)$. $\frac{2}{15}$
$B$. $P(\bar{B})$$(ii)$. $\frac{4}{15}$
$C$. $P(A \cap \bar{B})$$(iii)$. $\frac{8}{15}$
$D$. $P(B \cap \bar{A})$$(iv)$. $\frac{2}{3}$
$(v)$. $\frac{3}{7}$

It is given that $A$ and $B$ are such that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$. Then $P(B) = $?

If $80 \%$ of flights depart on time, $70 \%$ of flights arrive on time and $65 \%$ of flights depart on time and arrive on time, then the probability that a flight that has just departed on time will arrive on time is

If $\overline{E}$ and $\overline{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P(F) < 1$,then

Let $A$ and $B$ be two independent events such that $P(A) + P(B) = \frac{3}{4}$ and $P(\overline{A} | B) = \frac{2}{5}$. Then,$P(A \cap B)$ is -

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