$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probabilities of the two events are respectively:

  • A
    $\frac{1}{2}$ and $\frac{1}{3}$
  • B
    $\frac{1}{5}$ and $\frac{1}{6}$
  • C
    $\frac{1}{2}$ and $\frac{1}{6}$
  • D
    $\frac{2}{3}$ and $\frac{1}{4}$

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Two players,$P_1$ and $P_2$,play a game against each other. In every round,each player rolls a fair die once. Let $x$ and $y$ denote the outcomes for $P_1$ and $P_2$. If $x > y$,$P_1$ scores $5$ points and $P_2$ scores $0$. If $x = y$,each scores $2$ points. If $x < y$,$P_1$ scores $0$ and $P_2$ scores $5$. Let $X_n$ and $Y_n$ be the total scores of $P_1$ and $P_2$ after $n$ rounds. Match the following:
List-$I$ List-$II$
$(I)$ Probability of $(X_2 \geq Y_2)$ is $(P)$ $\frac{3}{8}$
$(II)$ Probability of $(X_2 > Y_2)$ is $(Q)$ $\frac{11}{16}$
$(III)$ Probability of $(X_3 = Y_3)$ is $(R)$ $\frac{5}{16}$
$(IV)$ Probability of $(X_3 > Y_3)$ is $(S)$ $\frac{355}{864}$
$(T)$ $\frac{77}{432}$

Let $B_{i} (i=1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occurs is $\alpha$,only $B_{2}$ occurs is $\beta$,and only $B_{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_{i}$ occurs,and these $4$ probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$ (All the probabilities are assumed to lie in the interval $(0, 1)$). Then $\frac{P(B_{1})}{P(B_{3})}$ is equal to ..........

In a game of throwing a die,what is the probability of getting a $1$ on an even-numbered throw?

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$A$ and $B$ each select one number at random from the distinct numbers $1, 2, 3, \ldots, n$. The probability that the number selected by $A$ is less than the number selected by $B$ is $\frac{1009}{2019}$. The probability that the number selected by $B$ is the number immediately next to the number selected by $A$ is:

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P(A)$ is equal to (Here, $\bar{E}$ is the complement of the event $E$)

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