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 ) $
$=0.05-0.02+0.10-0.02=0.11$
Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them problem
The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $
If $A$ and $B$ are two events, then the probability of the event that at most one of $A, B$ occurs, is
If $P(A) = 0.25,\,\,P(B) = 0.50$ and $P(A \cap B) = 0.14,$ then $P(A \cap \bar B)$ is equal to
If an integer is chosen at random from first $100$ positive integers, then the probability that the chosen number is a multiple of $4$ or $6$, is