Let a computer program generate only the digits $0$ and $1$ to form a string of binary numbers. The probability of occurrence of $0$ at even places is $\frac{1}{2}$ and the probability of occurrence of $0$ at odd places is $\frac{1}{3}$. Then the probability that $'10'$ is followed by $'01'$ is equal to:

Consider the system of equations $ax+by=0, cx+dy=0$,where $a, b, c, d \in \{0, 1\}$.
$STATEMENT-1$: The probability that the system of equations has a unique solution is $3/8$.
$STATEMENT-2$: The probability that the system of equations has a solution is $1$.

First bag contains $3$ red and $5$ black balls and second bag contains $6$ red and $4$ black balls. $A$ ball is drawn from each bag. The probability that one ball is red and the other is black,is

Let $A, B$ and $C$ be three events such that the probability that exactly one of $A$ and $B$ occurs is $(1-k)$,the probability that exactly one of $B$ and $C$ occurs is $(1-2k)$,the probability that exactly one of $C$ and $A$ occurs is $(1-k)$ and the probability that all $A, B$ and $C$ occur simultaneously is $k^2$,where $0 < k < 1$. Then the probability that at least one of $A, B$ and $C$ occurs is:

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25}$. If $P(T)$ denotes the probability of occurrence of the event $T$,then which of the following is true?
$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$

For three events $A, B$ and $C$,$P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. Then the probability that at least one of the events occurs is: