The probabilities of three events A, B and C aregiven by P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.If P(A$ \cup $B) = 0.8, P(A$ \cap $C) = 0.3, P(A$ \cap $B$ \cap $C) = 0.2,P(B$ \cap $C) = $\beta $ and P(A$ \cup $B$ \cup $C) = $\alpha $, where0.85 $ \le \alpha \le $ 0.95, then $\beta $ lies in the interval :

Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) = ${5 \over 9}$, is :

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

Let $N$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations $x + y + z = 1$ , $2x + Ny + 2z = 2$, $3x + 3y + Nz = 3$ . has a unique solution is $\dfrac{k}{6}$, then the sum of the value of $k$ and all possible values of $N$ is:

If $A$ and $B$ are any two events such that $P(A) = \dfrac{2}{5}$ and $P(A \cap B) = \dfrac{3}{20}$, then the conditional probability $P\big(A \mid (A' \cup B')\big)$, where $A'$ denotes the complement of $A$, is equal to:

A bag contains $6$ balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least $5$ black balls is: