Find $\mathbb P(A\cup B)$ given that $\mathbb P(A)=\frac15$ and $\mathbb P(B\mid A)=\frac7{20}$.
Here's a similar question, but I don't have the advantage of knowing $\mathbb P(B)$. Is it needed to find the solution to this exercise?
I know that
$$\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)$$
and
$$\mathbb P(A\cap B)=\mathbb P(B\mid A)\cdot\mathbb P(A)=\frac7{100}$$
However, I'm not sure how to compute $\mathbb P(B)$. I can show that
$$\begin{align*} \mathbb P(B)&=\mathbb P(B\cap A)+\mathbb P(B\cap A^C)\\[1ex] &=\frac7{100}+\mathbb P(B\mid A^C)\cdot\mathbb P(A^C)\\[1ex] &=\frac7{100}+\mathbb P(A^C\mid B)\cdot\mathbb P(B)\\[1ex] &=\frac7{100}+(1-\mathbb P(A\mid B))\cdot\mathbb P(B)\\[1ex] \implies\mathbb P(B)&=\frac7{100}\cdot\mathbb P(A\mid B) \end{align*}$$
but then I have the problem of finding this new conditional probability. I can't think of any other manipulations that don't lead in circles.
Is the question missing essential information to find the desired probability?
It's not determined.
Scenario $I$: you draw uniformly from $\{1,\cdots, 100\}$. $A$ is the event "the draw is $≤20$." $B$ is the event "the draw is $≤7$". Then $P(A\cup B)=P(A)=\frac 15$.
Scenario $II$: you draw uniformly from $\{1,\cdots, 100\}$. $A$ is the event "the draw is $≤20$". $B$ is the event "the draw is $≤7$ or $=21$". Then $P(A\cup B)=\frac {21}{100}$.