Probability Test 22

Let, $${ E }_{ 1 }$$ be the event noted number is $$7$$.
$${ E }_{ 2 }$$ be the event noted number is $$8$$.
$$H$$ be getting head on coin.
$$T$$ be getting tail on coin.

Let  $$\displaystyle E_{1}$$ be the event that a subject is selected from first group.
$$\displaystyle E_{2}$$ the event that a subject is selected from the second group.
$$E$$ be the event that an engineering subject is selected.
Now the probability that die shows $$3$$ or $$5$$  is
$$\displaystyle P(E_1)=\frac{2}{6}=\frac{1}{3}$$

Total number of bags $$=5$$
Let A be the group in which there are $$3$$ bags
and B be  the group in which there are $$2$$ bags

Now, probability of selecting bag from group A is $$\dfrac{3}{5}$$
probability of selecting bag from group B is $$\dfrac{2}{5}$$

Now, probability of selecting a black ball from first group is $$\dfrac{3}{5} \times \dfrac{2}{7}=\dfrac{6}{35}$$
Also, probability of selecting a black ball from second group is $$\dfrac{2}{5} \times \dfrac{4}{5}=\dfrac{8}{25}$$

Probability of first group , given it is black is $$=\dfrac{\dfrac{6}{35}}{\dfrac{6}{35}+\dfrac{8}{25}}=\dfrac{15}{43}$$

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 
$$ \displaystyle =\frac { P\left( S \right) P\left( \frac { F }{ S }  \right)  }{ P\left( S \right) \times P\left( \frac { F }{ S }  \right) +P\left( { S }^{ 1 } \right) \times P\left( \frac { E }{ { S }^{ 1 } }  \right)  }$$ 
$$ \displaystyle =\frac { \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 }  }{ \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } \times \dfrac { 1 }{ 4 }  } =\frac { 1 }{ 8 } \times \frac { 24 }{ 8 } =\frac { 3 }{ 8 }  $$