If $A$ and $B$ are two events such that $P(A | B) = 0.6$, $P(B | A) = 0.3$, and $P(A) = 0.1$, then $P(\bar{A} \cap \bar{B})$ equals:

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

$A$ die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of $4$. Then the conditional probability that the score $4$ has appeared at least once is

Consider three sets $E_1=\{1,2,3\}, F_1=\{1,3,4\}$ and $G_1=\{2,3,4,5\}$. Two elements are chosen at random,without replacement,from the set $E_1$,and let $S_1$ denote the set of these chosen elements.
Let $E_2=E_1-S_1$ and $F_2=F_1 \cup S_1$. Now two elements are chosen at random,without replacement,from the set $F_2$ and let $S_2$ denote the set of these chosen elements.
Let $G_2=G_1 \cup S_2$. Finally,two elements are chosen at random,without replacement,from the set $G_2$ and let $S_3$ denote the set of these chosen elements.
Let $E_3=E_2 \cup S_3$. Given that $E_1=E_3$,let $p$ be the conditional probability of the event $S_1=\{1,2\}$. Then the value of $p$ is

For a biased die,the probabilities for different faces to turn up are

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is

If $P(AB) = P(A)P(B)$,$P(A/B) = 1/4$,and $P(B/A) = 1/3$,then which of the following is true?