Question

I post a letter to my friend and do not receive a reply. It is known that one letter out of $m$ letters do not reach its destination. If it is certain that my friend will reply if he receives the letter. If $A$ denotes the event that my friend receives the letter and $B$ that I get a reply, then

• A. $P\left(B\right)=(1-1/m{)}^2$
• B. $P(A\cap B)=(1-1/m{)}^2$
• C. $P\left(A|B^{\prime }\right)=(m-1)/(2m-1)$
• D. $P(A\cup B)=(m-1)/m$

Answer 1

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

The correct options are
A $P\left(B\right)=(1-1/m{)}^2$
B $P\left(A|B^{\prime }\right)=(m-1)/(2m-1)$
C $P(A\cup B)=(m-1)/m$
D $P(A\cap B)=(1-1/m{)}^2$

$P\left(A\right)=\frac{m-1}{m},P\left(A^{\prime }\right)=\frac{1}{m}$
$P\left(B|A\right)=\frac{m-1}{m},P\left(B|A^{\prime }\right)=0$
Now $P\left(B|A\right)=\frac{m-1}{m}\Rightarrow \frac{P(A\cap B)}{P\left(A\right)}=\frac{m-1}{m}$
$\Rightarrow P(A\cap B)=\frac{(m-1{)}^2}{m^2}$
Also $P\left(B\right)=P\left(A\right)P\left(B|A\right)+P\left(A^{\prime }\right)P\left(B|A^{\prime }\right)$
$=\left(\frac{m-1}{m}\right)\left(\frac{m-1}{m}\right)+\left(\frac{1}{m}\right)\left(0\right)={\left(\frac{m-1}{m}\right)}^2$
$\Rightarrow P\left(B^{\prime }\right)=1-P\left(B\right)=1-\frac{(m-1{)}^2}{m^2}=\frac{2m-1}{m^2}$
$\therefore P\left(A|B^{\prime }\right)=\frac{P(A\cap B)}{P\left(B^{\prime }\right)}=\frac{P\left(A\right)-P(A\cap B)}{P\left(B^{\prime }\right)}$
$=\frac{m-1}{2m-1}$
$P(A\cup B)=\frac{m-1}{m}+{\left(\frac{m-1}{m}\right)}^2-{\left(\frac{m-1}{m}\right)}^2$.