# Let $A$ and $B$ be independent events such that $P(B|A\cup B) = \frac23$ and $P(A|B) =\frac12$. Find $P(B)$…

1. Let $A$ and $B$ be independent events such that $P(B|A\cup B) = \frac23$ and $P(A|B) = \frac12$. Find $P(B)$.

If $B$ is independent then it doesn't depend on the probability of $A\cup B$ so would $P(B)$ just be $\frac23$?

1. If $P(A) = \frac13$ and $P(B|A^c) = \frac14$, find $P(A\cup B)$.

$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$

First I tried to find $P(A\cap B)$,

$P(A\cap B) = P(A)P(B|A)$

$P(A) = \frac13, P(B|A) = 1-P(B|A^c) = \frac34$

$P(A\cap B) = \frac13 \frac34 = \frac14$

$P(A\cup B) = \frac13 + P(B) - \frac14$

To get $P(B)$,

$P(B)= P(B|A)⋅P(A)+P(B|A^C)⋅P(A^C)$

$P(B)= \frac34 \frac13 + \frac14 \frac23 = \frac5{12}$

$P(A\cup B) = \frac13 + \frac5{12} - \frac14 = \frac12$

Is this correct?

Guide:

(a)

$$\frac{P(B)}{P(A \cup B)}=\frac23$$

$$P(A)=\frac12$$

Hence $$P(A \cup B)=P(A)+P(B)-P(A)P(B)=\frac12+P(B)-\frac12P(B)=\frac12+\frac12P(B)$$

Substitute that into the first equation and you can solve for $P(B)$.

(b)

Use

$$P(A \cup B) = P(A^C \cap B)+P(A)$$

$$P(B \cap (A \cup B))=P(B),$$
and since $P(A \cup B) < 1$, hence $B$ and $A \cup B$ are not independent.
$$P(B|A)=1-P(B|A^c)$$ is not true in general.
We do have $$P(B|A)=1-P(B^c|A)$$ though.