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

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 exactly one of them solves the problem is given by,

$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A})$

$=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}$

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$

Let $A$ and $B$ be events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals

One card is drawn at random from a well shuffled deck of $52$ cards. In which of the following cases are the events $\mathrm{E}$ and $\mathrm{F}$ independent ?

$E:$ 'the card drawn is a spade'

$F:$ 'the card drawn is an ace'

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$

An integer is chosen at random from the integers $\{1,2,3, \ldots \ldots . .50\}$. The probability that the chosen integer is a multiple of atleast one of $4,6$ and $7$ is

- [JEE MAIN 2024]

If $A$ and $B$ are two events such that $P\,(A \cup B)\, + P\,(A \cap B) = \frac{7}{8}$ and $P\,(A) = 2\,P\,(B),$ then $P\,(A) = $