Conditional Probability and Multiplication Law

The tree diagram given below shows the outcomes for Hildy choosing an outfit from 3 different pairs of pants and 3 different shirts. Assuming there is an equal chance of Hildy choosing from any of the pants and shirts. What is the probability that Hildy will choose the blue pants and red shirt?

If $A\text{ and }B$ are two disjoint events associated with the sample space $S$ and $F$ is any event of $S$ such that $P\left(F\right)\ne 0$, then find $P\left(\right(A\cup B)\mid F)$.

The tree diagram given below shows the outcomes for Hildy choosing an outfit from 3 different pairs of pants and 3 different shirts. Assuming there is an equal chance of Hildy choosing from any of the pants and shirts. Then prove that the probability that Hildy will choose the blue pants and red shirt is $\frac{1}{9}$.

If $A,B$ are two independent events and $P\left(A\right)=0.4$ and $P\left(A\cap B\right)=0.1$, find $P\left(B\right)$.

If $A,B$ are two independent events and $P\left(A\right)=0.4$ and $P\left(A\cap B\right)=0.2$, find $P\left(B\right)$.

If $A\text{ and }B$ are any two events associated with the sample space $S$ and $F$ is any event of $S$ such that $P\left(F\right)\ne 0$, then find $P\left(\right(A\cup B)\mid F)$.

An experiment has $10$ equally likely outcomes: three are favourable to event $A,$ five are favourable to event $B$ and four are favourable to neither $A$ nor $B$ Find $P(A\mid B)$ and $P(B\mid A).$

$A$ speaks truth in $75\%$ of the cases, while $B$ in $90\%$ of the cases. In what percent of cases, are they likely to contradict each other in stating the same fact?

A fair die is rolled. Consider events $A=\left\{1,3,5\right\}, B=\left\{2,3\right\}$ and $C=\left\{2,3,4,5\right\}$, then find $P\left(\frac{A\cup B}{C}\right)$ and $P\left(\frac{A\cap B}{C}\right)$.