Question

Let $A$ and $B$ be events such that $P\left(\overline{A}\right)=\frac{4}{5},P\left(B\right)=\frac{1}{3},P\left(\frac{A}{B}\right)=\frac{1}{6}$, then:

• A. $P(A\cap B)$
• B. $P(A\cup B)$
• C. $P\left(\frac{B}{A}\right)$
• D. Are $A$ and $B$ independent

Answer 1

The correct option is B $P(A\cup B)$

$P\left(\overline{A}\right)=\frac{4}{5}$ then $P\left(A\right)=\frac{1}{5}$

$P\left(B\right)=\frac{1}{3}$ and $P\left(\frac{A}{B}\right)=\frac{1}{6}$

we know that $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$

$\frac{1}{6}=\frac{A(A\cap B)}{1}\cdot 3$

$P(A\cap B)=\frac{1}{1^8}$

we know that $P\left(A\right)+P\left(B\right)=P(A\cup B)+P(A\cap B)$

$\frac{1}{5}+\frac{1}{3}-\frac{1}{18}=P(A\cup B))$

$P(A\cup B)=\frac{18-5+30}{90}=\frac{43}{90}$

$P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)}=\frac{1\cdot 5}{18\cdot 1}=\frac{5}{18}$