Probability Test 23

Let $$E$$ be the event that employee received the letter and $$A$$ that employer received the reply, then
$$\displaystyle P\left ( E \right )= \frac{n-1}{n}$$ and $$\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$$

$$\displaystyle \sum_{k = 0}^{k = n}P(E_k) = 1$$
$$\therefore \displaystyle \sum_{k = 0}^{k = n}\alpha k = 1$$

$$P(E_k)=C$$ for $$0\leq k\leq n$$ and $$\displaystyle \sum_{k=0}^nP(E_k)=1\Rightarrow C=1/(n+1)$$.
Also $$\displaystyle P(A|E_k)=\frac {^{n-1}C_{k-1}}{^nC_k}=\frac {k}{n}$$
By the total probability rule
$$P(A)=\displaystyle \sum_{k=0}^nP(E_k)P(A|E_k)$$
$$\displaystyle =\frac {1}{n(n+1)}\frac {n(n+1)}{2}=\frac {1}{2}$$.

Given $$P(E_k)\propto k$$ for $$0 \le k \le n$$

Let $$E_1, E_2$$ and $$A$$ denote the following events:
$$E_1$$: coin selected is fair
$$E_2$$: coin selected is biased
$$A: $$ the first toss results in a head and the second toss results in a tail.
$$\displaystyle P(E_1)=\frac {m}{N}, P(E_2)=\frac {N-m}{N}$$,
$$\displaystyle P(A|E_1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E_2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$$.
By Bayes' rule
$$\displaystyle P(E_1|A)=\frac {P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\frac {9m}{8N+m}$$.

Let $$P(E_k)=k\alpha$$ where $$\alpha$$ is the constant of proportionality.
Since $$\displaystyle \sum_{k=0}^nP(E_k)=1$$, we get $$\alpha=2/n(n+1)$$
Now, $$P(A)=\displaystyle \sum_{k=0}^nP(E_k)P(A|E_k)=\frac {2}{n^2(n+1)}\sum_{k=0}^nk^2$$
$$\displaystyle =\frac {2}{n^2(n+1)}\frac {n(n+1)(2n+1)}{6}=\frac {2n+1}{3n}$$

Let $$B=$$ {exactly one of $${S}_{1}$$ and $${S}_{2}$$ is among of the six winners}.

Let $${ A }_{ 1 }$$ denote the event that a coin having heads on both sides is chosen, and $${ A }_{ 2 }$$ denote the vent that a fiar coin is chosen.
Let $$E$$ denote the vent that head occurs, Then
$$\displaystyle P\left( { A }_{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A }_{ 2 } \right) =\frac { n+1 }{ 2n+1 } $$
Probability of occurrence of event $$E$$, if unfair coin was selected is $$\displaystyle P\left( \frac { E }{ { A }_{ 1 } }  \right) =1$$
Probability of occurrence of event $$E$$, if fair coin was selected is $$\displaystyle P\left( \frac { E }{ { A }_{ 2 } }  \right) =\frac { 1 }{ 2 } $$
$$\because P\left( E \right) =P\left( { A }_{ 1 }\cap E \right) +P\left( { A }_{ 2 }\cap E \right) $$
$$\displaystyle \therefore P\left( E \right) =P\left( { A }_{ 1 } \right) P\left( \frac { E }{ { A }_{ 1 } }  \right) +P\left( { A }_{ 2 } \right) P\left( \frac { E }{ { A }_{ 2 } }  \right) $$
$$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right)  } \\ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$$

Given,The probability of a Jee Aspirant to be successful if he studies for 10 hours per day,$$P(t_{10})=0.8$$ 
The probability of a Jee Aspirant to be successful if he studies for 7 hours per day,$$P(t_{7})=0.6$$
The probability of a Jee Aspirant to be successful if he studies for 4 hours per day,$$P(t_{4})=0.4$$
The probability that she will study 10 hours per day,$$\displaystyle P(A)=0.1$$
The probability that she will study 7 hours per day,$$\displaystyle P (B)=0.2$$
The probability that she will study 4 hours per day,$$\displaystyle P(C)=0.7$$
$$\therefore The\;probability\;that\;she\;will\;be\;successful,P(S)=P(t_{10})*P(A)+P(t_{7})*P(B)+P(t_{4})*P(S/t_{4})$$
$$=0.8*0.1+0.6*0.2+0.4*0.7=0.48$$
But probability that she studies 4 hours per day for success=$$P(t_{4} \cap C)=P(t_{4})P(C)=0.4*0.7=0.28$$
Since $$(A,t_{10}),(B,t_{7}),(C,t_{4})$$ pairwise independent $$\Rightarrow P(A \cap B)=P(A).P(B)$$
$$\therefore$$ The probability that she studies 4 hours per day given she is successful,$$P(C/S)=\displaystyle\frac{P(C \cap S)}{P(S)}$$
$$=\displaystyle\frac{0.28}{0.48}=\displaystyle\frac{7}{12}$$