If $E_1$ and $E_2$ are two events of the sample space such that $P(E_1) = \frac{1}{4}$,$P(E_1 | E_2) = \frac{1}{2}$ and $P(E_2 | E_1) = \frac{1}{3}$,then $P(E_1 | \bar{E}_2) = $

Let $X$ be the set of all five-digit numbers formed using $1, 2, 2, 2, 4, 4, 0$. For example,$22240$ is in $X$ while $02244$ and $44422$ are not in $X$. Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of $20$ given that it is a multiple of $5$. Then the value of $38p$ is equal to

$A$ parent has two children. If at least one of them is a boy,then the probability that the other is also a boy is:

Let $S$ be the sample space of all $3 \times 3$ matrices with entries from the set $\{0, 1\}$. Let the events $E_1$ and $E_2$ be given by $E_1 = \{A \in S : \operatorname{det} A = 0\}$ and $E_2 = \{A \in S : \text{sum of entries of } A \text{ is } 7\}$. If a matrix is chosen at random from $S$,then the conditional probability $P(E_1 \mid E_2)$ equals:

Two dice are thrown and the sum of the numbers appearing on the dice is observed to be a multiple of $4$. If $p$ is the conditional probability that number $4$ has appeared at least once,then $3p + 2 =$

Suppose $E$ and $F$ are two events of a random experiment. If the probability of occurrence of $E$ is $1/5$ and the probability of occurrence of $F$ given $E$ is $1/10$,then the probability of non-occurrence of at least one of the events $E$ and $F$ is