Probability Test - 49

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 $$\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$$

Let $$A$$ denote the event that the sample contains exactly three white balls and let $$B$$ be the event that the ball
Now $$\displaystyle p\left(A \right)=\left ( ^{4}c_{3} \right )\frac{8}{12}\times\frac{8}{12}\times\frac{8}{12}\times\frac{4}{12}$$
 $$\displaystyle P\left(AB \right)=\left ( ^{3}C_{2} \right )\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}$$ 
$$\displaystyle \therefore P\left [ B/A \right ]=\frac{P\left [ AB \right ]}{P\left ( A \right )}=\frac{^{3}C_{2}}{^{4}C_{2}}=\frac{3}{4}$$
In the case of sampling without replacement,
$$\displaystyle P\left(A \right)=\left ( ^{4}C_{3} \right )\frac{8}{12}\times\frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$$
 $$\displaystyle P\left (AB \right )=\left ( ^{3}C_{2} \right )\frac{8}{12}\times\frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$$ 
$$\displaystyle \therefore  P\left (AB \right )=\frac{^{3}C_{2}}{^{4}C_{3}}=\frac{3}{4}$$

$$\displaystyle P(E'\cap F'|G)=\frac {P(E'\cap F'\cap G)}{P(G)}$$
$$\displaystyle =\frac {P(G)-P[(E\cap G)\cup (F\cap G)]}{P(G)}$$
$$\displaystyle =\frac {P(G)-P(E\cap G)-P(F\cap G)}{P(G)}=\frac {P(G)[1-P(E)-P(F)]}{P(G)}$$
$$=P(E')-P(F)$$.

$$E_1$$: biased coin is chosen,
$$E_2$$: fair coin is choice.
and A: toss results in a head.
$$P(E_1)=n/(2n+1), P(E_2)=(n+1)/(2n+1)$$
$$P(A|E_1)=1$$ and $$P(A|E_2)=1/2$$
By the total probability rule
$$\dfrac {31}{42}=\dfrac {n}{2n+1}(1)+\dfrac {(n+1)}{2(2n+1)}\Rightarrow \dfrac {3n+1}{2n+1}=\dfrac {31}{21}\Rightarrow n=10$$

Let $$E_1(0\leq i\leq 2)$$ denote the event that urn contains $$i$$ white and $$(2-i)$$ black balls.
Let $$A$$ denote the event that a white ball is drawn from the urn.
We have $$P(E_i)=1/3$$ for $$i=0, 1, 2$$. and $$P(A|E_1)=1/3, P(A|E_2)=2/3, P(A|E_3)=1$$.
By the total probability rule,
$$P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)$$
$$\displaystyle =\frac {1}{3}\left [\frac {1}{3}+\frac {2}{3}+1\right ]=\frac {2}{3}$$

Let $$E_i$$ denote the event that the bag contains $$i$$ black and ($$10-i$$) white balls $$(i=0, 1, 2, ...., 10)$$. Let $$A$$ denote the event that the three balls drawn at random from the bag are black. We have
$$P(E_i)=\dfrac {1}{11} (i=0, 1, 2, ...., 10)$$
$$P(A|E_i)=0$$ for $$i=0, 1, 2$$
and $$P(A|E_i)=\dfrac {^iC_3}{^{10}C_3}$$ for $$i\geq 3$$
Now, by the total probability rule
$$\displaystyle P(A)=\sum_{i=0}^{10}P(E_i)P(A|E_i)$$
$$=\frac {1}{11}\times \frac {1}{^{10}C_3}[^3C_3+^4C_4+....+^{10}C_3]$$
But $$^3C_3+^4C_3+^5C_3+....+^{10}C_3$$
$$=^4C_4+^4C_3+^5C_3+...+^{10}C_3$$
$$=^5C_4+^5C_3+^6C_3+....+^{10}C_3$$
$$=^6C_4+^6C_3+....+^{10}C_3=....=^{11}C_4$$
Thus, $$P(A)=\dfrac {^{11}C_4}{11\times ^{10}C_3}=\dfrac {1}{4}$$
By the Bayes' rule
$$P(E_9|A)=\dfrac {P(E_9)P(A|E_9)}{P(A)}=\dfrac {\dfrac {1}{11}\dfrac {(^9C_3)}{^{10}C_3}}{\dfrac {1}{4}}=\dfrac {14}{55}$$.

Let $$E_i$$ denote the event that the bag contains $$ i $$ black and $$ (10-i)$$  white balls $$(i=0, 1, 2, ..., 10)$$. 

Let $$E_1$$ denote the event that the letter came from $$TATANAGAR$$ and
$$E_2$$ the event that the letter came from $$CALCUTTA$$, Let $$A$$ denote the event that the two consecutive alphabets visible on the envelops are $$TA$$.
We have $$P(E_1)=1/2, P(E_2)=1/2, P(A|E_1)=2/8, P(A|E_2)=1/7$$.
Therefore, by Baye's theorem we have
$$\displaystyle P(E_2|A)=\frac {P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}=\frac {4}{11}$$.

Let $$E$$ be the event that the man reports that six occurs whle throwing the die and let $$S$$ be the event that six occurs. Then 
$$P(S)=$$ Probability that six occurs $$ \displaystyle =\frac { 1 }{ 6 }   $$
$$P\left( { S }^{ 1 } \right) =$$ probability that six does not occur $$ \displaystyle =1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } $$
$$ \displaystyle P\left( \frac { F }{ S }  \right) = $$ probability that the man reports that six occurs when six has actually occurred 
$$=$$ probability that the man reports the truth $$ \displaystyle =\frac { 3 }{ 4 }  $$ 
$$ \displaystyle P\left( \frac { E }{ { S }^{ 1 } }  \right) =$$ probability that the man report that six occur when six has not actually occurred. 
$$=$$ probability that the man does not speak the truth 
$$ \displaystyle 1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } . $$ 
By Bayes' theorem 
$$ \displaystyle P\left( \frac { S }{ E }  \right) =$$ probability that the man 
reports that six occurs when six has actually occured 
$$ \displaystyle =\frac { P\left( S \right) P\left( \frac { F }{ S }  \right)  }{ P\left( S \right) \times P\left( \frac { F }{ S }  \right) +P\left( { S }^{ 1 } \right) \times P\left( \frac { E }{ { S }^{ 1 } }  \right)  }$$ 
$$ \displaystyle =\frac { \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 }  }{ \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } \times \dfrac { 1 }{ 4 }  } =\frac { 1 }{ 8 } \times \frac { 24 }{ 8 } =\frac { 3 }{ 8 }  $$

We wish to find the probability that $$X\geq 6$$, given that $$X>3$$. 
We know that the first three tosses have been unsuccessful. 
Hence, 
$$P\left(X \geq 6 | X>3 \right) = 1 - P(X = 4 | x >3 ) - P(X =5 |X > 3) $$
$$\Rightarrow P\left(X \geq 6 | X>3 \right) = 1 - \dfrac{1}{6} - \dfrac{1}{6} \times \dfrac56 $$
$$\Rightarrow P\left(X \geq 6 | X>3 \right) = 1 - \dfrac{11}{36}$$
$$\Rightarrow P\left(X \geq 6 | X>3 \right) =  \dfrac{25}{36}$$