Question

A, B are two events with positive probabilities. If $P\left(A\right)=0.4$, $P\left(B/A\right)=0.9$, $P\left(\overline{B}/\overline{A}\right)=0.6$ then

• A. $P(A\cup B)=0.64$
• B. $P(A\cap B)=0.36$
• C. $P\left(A/B\right)=0.6$
• D. $P\left(A/\overline{B}\right)=0.25$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $P(A\cup B)=0.64$
B $P(A\cap B)=0.36$
C $P\left(A/B\right)=0.6$

$P\left(B/A\right)=0.9\Rightarrow \frac{P(A\cap B)}{P\left(A\right)}=0.9\Rightarrow P(A\cap B)=0.36$

$P\left(\overline{B}/\overline{A}\right)=0.6\Rightarrow \frac{P(\overline{B}\cap \overline{A})}{P\left(\overline{A}\right)}=0.6\Rightarrow \frac{P\left(\overline{A\cup B}\right)}{P\left(\overline{A}\right)}=0.6\Rightarrow \frac{1-P(A\cup B)}{1-P\left(A\right)}=0.6\Rightarrow \frac{1-(A\cup B)}{1-\left(0.4\right)}=0.6\Rightarrow A\cup B=0.64$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow P\left(B\right)=0.64+0.36-0.4=0.6\therefore P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}\Rightarrow P\left(A/B\right)=\frac{0.36}{0.6}=0.6$

$P\left(A/\overline{B}\right)=\frac{P(A\cap \overline{B})}{P\left(\overline{B}\right)}\Rightarrow P\left(A/\overline{B}\right)=\frac{P\left(A\right)-P(A\cap B)}{1-P\left(B\right)}\Rightarrow P\left(A/\overline{B}\right)=\frac{0.4-0.36}{1-0.6}=0.1$