Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10 .$ The probability that both will qualify the examination is $0.02 .$ Find the probability that Only one of them will qualify the examination.
Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination, respectively. Given that
$P(E)=0.05$, $P(F)=0.10$ and $P(E \cap F)=0.02$
Then
The event only one of them will qualify the examination is same as the event either (Anil will qualify, andAshima will not qualify) or (Anil will not qualify and Ashima will qualify) i.e., $E \cap F ^{\prime}$ or $E ^{\prime} \cap F ,$ where $E \cap F ^{\prime}$ and $E ^{\prime} \cap F$ are mutually exclusive.
Therefore, $P$ (only one of them will qualify) $=P(E \cap F^{\prime} $ or $E^{\prime} \cap F)$
$= P \left( E \cap F ^{\prime}\right)$ $+ P \left( E ^{\prime} \cap F \right)$ $= P ( E )- P ( E \cap F )+ P ( F )- P ( E \cap F ) $
$=0.05-0.02+0.10-0.02=0.11$
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
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
Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=p .$ Find $p$ if they are mutually exclusive.
If two events $A$ and $B$ are such that $P\,(A + B) = \frac{5}{6},$ $P\,(AB) = \frac{1}{3}\,$ and $P\,(\bar A) = \frac{1}{2},$ then the events $A$ and $B$ are
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 $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$