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 the problem is solved.
Probability of solving the problem by $\mathrm{A}, \mathrm{P}(\mathrm{A})=\frac{1}{2}$
Probability of solving the problem by $\mathrm{B}, \mathrm{P}(\mathrm{B})=\frac{1}{3}$
since the problem is solved independently by $A$ and $B$,
$\therefore $ $\mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$
$P(A^{\prime})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$
$P(B^{\prime})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$
Probability that the problem is solved $=\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AB})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$=\frac{4}{6}$
$=\frac{2}{3}$
If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A \cap B)$ if $A$ and $B$ are independent events
If $A$ and $B$ are any two events, then the probability that exactly one of them occur is
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 $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
In class $XI$ of a school $40\%$ of the students study Mathematics and $30 \%$ study Biology. $10 \%$ of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$, then the incorrect statement amongst the following statements is