Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

If $A$ and $B$ are two events,then the probability of the event that at most one of $A$ and $B$ occurs is:

Let $A, B, C$ be three pairwise independent events of a random experiment. If $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$, $P(A) > 0$, $P(B) = b$, and $P(C) = c$, then $P((\bar{B} \cap \bar{C}) \mid A) = $

$A$ computer program has two modules $X$ and $Y$ and errors in them occur independently. $X$ has an error with probability $0.1$ and $Y$ has an error with probability $0.3$. If an error in $X$ alone causes the program to crash with probability $0.5$, an error in $Y$ alone causes the program to crash with probability $0.7$, and an error in both $X$ and $Y$ causes the program to crash with probability $0.8$, then the probability that the program crashes is

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2}$. Then a value of $\frac{P(E)}{P(F)}$ is

$E_1$ and $E_2$ are two independent events of a random experiment with $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$. Match the items of List-$I$ with those of List-$II$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 | E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 | E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$