Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$
Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
Statement $1$ is false, Statement $2$ is true.
Statement $1$ is true, Statement $2$ is true,Statement $2$ is not a correct explanation of Statement $1$ .
Statement $1$ is true, Statement $2$ is false
Statement $1$ is true, Statement $2$ is true;Statement $2$ is a correct explanation of Statement $1.$
If $P(A) = P(B) = x$ and $P(A \cap B) = P(A' \cap B') = \frac{1}{3}$, then $x = $
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( B \cap A ^{\prime}\right)$.
Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is
Let $A$ and $B$ be two events such that $P\,(A) = 0.3$ and $P\,(A \cup B) = 0.8$. If $A$ and $B$ are independent events, then $P(B) = $
The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be