$A$ number $n$ is chosen at random from $\{1, 2, 3, 4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder $1$ when divided by $7$ is:

If $P(A)=0.59, P(B)=0.30$ and $P(A \cap B)=0.21$,then $P(A^{\prime} \cap B^{\prime})$ is equal to

$A$ student appears for tests $I, II$ and $III$. The student is successful if he passes either in tests $I$ and $II$ or tests $I$ and $III$. The probabilities of the student passing in tests $I, II, III$ are $p, q$ and $\frac{1}{2}$ respectively. If the probability that the student is successful is $\frac{1}{2}$,then

When a certain biased die is rolled,a particular face occurs with probability $\frac{1}{6}-x$ and its opposite face occurs with probability $\frac{1}{6}+x$. All other faces occur with probability $\frac{1}{6}$. Note that opposite faces sum to $7$ in any die. If $0 < x < \frac{1}{6}$,and the probability of obtaining a total sum of $7$ when such a die is rolled twice is $\frac{13}{96}$,then the value of $x$ is:

Three coins are tossed once. Find the probability of getting $2$ heads.

Let $A = \{1, 3, 5, 7, 9\}$ and $B = \{2, 4, 6, 8\}$. If an ordered pair $(a, b)$ is chosen at random from the Cartesian product $A \times B$,what is the probability that $a + b = 9$?