Question

$A$ post a letter to $B$ and inform him over the telephone. It is known that one letter out of $10$ letters does not reach its destination. If it is certain that $B$ will reply if he receives the letter and even if $B$ does not receive the letter, he replies with a probability of $\frac{1}{5}$ as he got the information that $A$ has written him a letter. If $X$ denotes the event that $B$ receives the letter and $Y$ denotes the event that $A$ gets a reply, then

• A. $P(X\cap Y)=\frac{81}{100}$
• B. $P\left(X|\overline{Y}\right)=\frac{45}{86}$
• C. $A$ gets a reply, then the probability that $B$ receives the letter is $\frac{45}{46}$
• D. $A$ gets a reply, then the probability that $B$ receives the letter is $\frac{5}{46}$

Answer 1

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

$A$ gets a reply, then the probability that $B$ receives the letter is $\frac{45}{46}$

$P\left(X\right)=\frac{9}{10}$
$Y:$ event that $A$ gets a reply
$\Rightarrow P\left(Y\right)=\frac{9}{10}\times \frac{9}{10}+\frac{1}{10}\times \frac{1}{5}\times \frac{9}{10}$
$\Rightarrow P\left(Y\right)=\frac{81}{100}+\frac{9}{500}=\frac{414}{500}$


$P(X\cap Y)=$ Probability that $B$ receives letter and $A$ gets reply
$\Rightarrow P(X\cap Y)=\frac{9}{10}\times \frac{9}{10}=\frac{81}{100}$


$P\left(X|\overline{Y}\right)=$ Probability that $B$ receives the letter given that $A$ does not get reply
$\Rightarrow P\left(X|\overline{Y}\right)=\frac{P(X\cap \overline{Y})}{P\left(\overline{Y}\right)}$
$\Rightarrow P\left(X|\overline{Y}\right)=\frac{\frac{9}{10}-\frac{81}{100}}{1-\frac{414}{500}}$
$\therefore P\left(X|\overline{Y}\right)=\frac{45}{86}$


Probability that $B$ receives the letter given that $A$ gets reply
$P\left(X|Y\right)=\frac{P(X\cap Y)}{P\left(Y\right)}$
$\Rightarrow P\left(X|Y\right)=\frac{81/100}{414/500}$
$\therefore P\left(X|Y\right)=\frac{45}{46}$