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
It is said that $A$ denotes the event that my friend receives the letter and $B$ that I get the reply then:
This question has multiple correct options.
A. $P(B) = {\left( {1 - \dfrac{1}{m}} \right)^2}$
B. $P(A \cap B) = {\left( {1 - \dfrac{1}{m}} \right)^2}$
C. $P(A|B') = \dfrac{{m - 1}}{{2m - 1}}$
D. $P(A \cup B) = \dfrac{{m - 1}}{m}$

Answer 1

$A$ denotes the event that my friend receives the letter then $P(A) = \dfrac{{m - 1}}{m}$ because $m - 1$ are the letters that are received by the friend and one is not received by him then we can find the value of $P\left( {\dfrac{B}{A}} \right)$.

Complete step by step solution:
In this question it is given that 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. This means that the total of $m$ letters were post but did not get one letter so I could get only $m - 1$ letters
As $A$ is the probability that I get the letter received and there are $m - 1$ letters received out of the total of $m$ letters then we can say that
$P(A) = \dfrac{{m - 1}}{m}$
Now we know that for any event $X$
$P(X) = 1 - P(X')$
So we can say that for the above event also that
$\Rightarrow P(A) = 1 - P(A')$
$\Rightarrow P(A') = 1 - P(A) = 1 - \dfrac{{m - 1}}{m} = \dfrac{1}{m}$
And $P\left( {\dfrac{B}{A}} \right)$ is the probability of the event $B$ when $A$ event has already occurred.
$P\left( {\dfrac{B}{A}} \right) = \dfrac{{m - 1}}{m}$
And we know the formula that
$\Rightarrow P\left( {\dfrac{B}{A}} \right) = \dfrac{{P(A \cap B)}}{{P(A)}}$
$\Rightarrow \dfrac{{P(A \cap B)}}{{P(A)}} = \dfrac{{m - 1}}{m}$
$\Rightarrow P(A \cap B) = \dfrac{{m - 1(m - 1)}}{{m(m)}} = {\left( {1 - \dfrac{1}{m}} \right)^2}$
Now as we know that if the friend do not receive the letter I will not get the reply
So $P\left( {\dfrac{B}{{A'}}} \right) = 0$
Now know the formula that
$\Rightarrow P(B) = P(A)P\left( {\dfrac{B}{A}} \right) + P(A')P\left( {\dfrac{B}{{A'}}} \right)$
$\Rightarrow P(B) = \dfrac{{m - 1}}{m}.\dfrac{{m - 1}}{m} + \dfrac{1}{m}(0) =$ ${\left( {1 - \dfrac{1}{m}} \right)^2}$

$\Rightarrow P(B') = 1 - P(B) = 1 - \dfrac{{{{(m - 1)}^2}}}{{{{(m)}^2}}} = \dfrac{{2m - 1}}{{{m^2}}}$
$\Rightarrow P\left( {\dfrac{A}{{B'}}} \right) = \dfrac{{P(A \cap B')}}{{P(B')}}$
We know that $P(A \cap B') = P(A) - P(A \cap B)$
$\Rightarrow P\left( {\dfrac{A}{{B'}}} \right) = \dfrac{{\dfrac{{m - 1}}{m} - {{\left( {\dfrac{{m - 1}}{m}} \right)}^2}}}{{\dfrac{{2m - 1}}{{{m^2}}}}} = \dfrac{{m - 1}}{{2m - 1}}$
Also we know the formula that
$P(A \cup B) = P(A) + P(B) - P(A \cup B)$
$\Rightarrow {\left( {\dfrac{{m - 1}}{m}} \right)^2} = \left( {\dfrac{{m - 1}}{m}} \right) + {\left( {\dfrac{{m - 1}}{m}} \right)^2} - P(A \cup B) \\\ \Rightarrow P(A \cup B) = \left( {\dfrac{{m - 1}}{m}} \right) \\\$
So here we get that
$\Rightarrow P(A \cap B) = {\left( {1 - \dfrac{1}{m}} \right)^2}$
$\Rightarrow P(B) =$ ${\left( {1 - \dfrac{1}{m}} \right)^2}$
$\Rightarrow P\left( {\dfrac{A}{{B'}}} \right) = \dfrac{{m - 1}}{{2m - 1}}$
$\Rightarrow P(A \cup B) = \dfrac{{m - 1}}{m}$

So all the options A, B, C, D are correct.

Note:
you must know the formula of probability that
$\Rightarrow P(A \cup B) = P(A) + P(B) - P(A \cup B)$
$\Rightarrow P\left( {\dfrac{A}{{B'}}} \right) = \dfrac{{P(A \cap B')}}{{P(B')}}$
$\Rightarrow P(B') = 1 - P(B)$
Also we must know that $0 \leqslant P(A) \leqslant 1$.