Question

Mohan posts a letter to Sohan. It is known that one letter out of $10$ letters does not reach its destination. It is certain that Sohan will reply if he receives the letter. If $A$ denotes the event that Sohan receives the letter and $B$ denotes the event that Mohan gets a reply, then

• A. $P\left(B\right)=\frac{81}{100}$
• B. $P(A\cap B)=\frac{81}{100}$
• C. $P\left(A|\overline{B}\right)=\frac{9}{19}$
• D. $P(A\cup B)=\frac{9}{10}$

Answer 1

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

The correct options are
A $P\left(B\right)=\frac{81}{100}$
B $P(A\cap B)=\frac{81}{100}$
C $P\left(A|\overline{B}\right)=\frac{9}{19}$
D $P(A\cup B)=\frac{9}{10}$

$A:$ Sohan receives the letter.
$B:$ Mohan gets a reply.

$P\left(B\right)=P\left(\text{Sohan receives the letter}\right)\times P\text{(Sohan writes a reply)}\times P\left(\text{Mohan gets the letter}\right)$

$=\frac{9}{10}\times 1\times \frac{9}{10}=\frac{81}{100}$

$P(A\cap B)=P\left(B\right)=\frac{81}{100}$

$P\left(A|\overline{B}\right)=\frac{P(A\cap \overline{B})}{P\left(\overline{B}\right)}$
$P\left(\overline{B}\right)=1-P\left(B\right)=\frac{19}{100}\Rightarrow P(A\cap \overline{B})=\frac{9}{10}\times 1\times \frac{1}{10}\Rightarrow P\left(A|\overline{B}\right)=\frac{9}{19}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\Rightarrow P(A\cup B)=\frac{9}{10}+\frac{81}{100}-\frac{81}{100}$
$\Rightarrow P\left(A\right)=\frac{9}{10}$