Probability Test 7

C is correct

A: number of ace is drawn in the first n draw
B: an ace appear in the $${ \left( n+1 \right)  }^{ th }$$ draw
Hence the probability that exactly n cards are dealt before the first ace appear is equal to $$P\left( A\cap B \right) $$
$$\displaystyle P\left( A \right) =\frac { ^{ 48 }{ { C }_{ n } } }{ ^{ 52 }{ { C }_{ n } } } ,P\left( \frac { B }{ A }  \right) =\frac { 4 }{ 52-n } $$
$$\displaystyle \therefore P\left( A\cap B \right) =P\left( A \right) .P\left( \frac { B }{ A }  \right) =\frac { ^{ 48 }{ { C }_{ n } } }{ ^{ 52 }{ { C }_{ n } } } .\frac { 4 }{ 52-n } $$
$$\displaystyle =\frac { 48! }{ \left( 48-n \right) !n! } \times \frac { \left( 52-n \right) !n! }{ 52! } \times \frac { 4 }{ 52-n } $$
$$\displaystyle =\frac { 4\left( 51-n \right) \left( 50-n \right) \left( 49-n \right)  }{ 52\times 51\times 50\times 49 } $$

Let $$\displaystyle { E }_{ i }\left( 0\le i\le 2 \right) $$ denotes the event that urn contains $$i$$ white and $$2-i$$ black balls.

$${\textbf{Step 1: Consider all the possible events.}}$$

               $${\text{Let }}{E_1},{E_2}{\text{ and }}A{\text{ be the following events}}{\text{.}}$$

               $${E_1} = {\text{six occurs}}$$

               $$\therefore P\left( {{E_1}} \right) = \dfrac{1}{6}$$

               $${E_2} = {\text{six does not occurs}}$$

               $$\therefore P\left( {{E_1}} \right) = \dfrac{5}{6}$$

               $$A = {\text{Man reports that it is a six}}$$

               $${\text{Now, probability that the man reports that there is a six on the die is }}P\left( {A/{E_1}} \right)$$

               $$P\left( {A/{E_1}} \right) = \dfrac{3}{4}$$   $$\left( {{\textbf{Given}}} \right)$$

               $${\text{So, probability that the man reports that there is a six on the die }}$$ 

               $${\text{given that six has not ocured on the die }}P\left( {A/{E_2}} \right)$$

               $$ = {\text{ probability that man does not speak the truth}}$$

               $$ = 1 - \dfrac{3}{4}$$

               $$P\left( {A/{E_2}} \right) = \dfrac{1}{4}$$

$${\textbf{Step 2: Find the probability that it is actually a six.}}$$

               $${\text{The probability of the man speaking truth given the die shows the value 6, }}$$  

               $${\text{which is }}P{\text{ }}\left( {{E_1}|A} \right){\text{ which is given by, using Bayes theorem:}}$$

               $$ \Rightarrow P({E_1}|{\text{ }}A) = \dfrac{{P(A|{\text{ }}{E_1})P({E_1})}}{{P(A|{\text{ }}{E_2})P({E_2}) + P(A|{\text{ }}{E_1})P({E_1})}}$$         

               $$\left(\mathbf {{\text{By Bayes theorem: }}P(A|{\text{ }}B) = \dfrac{{P(B|{\text{ }}A)P(A)}}{{P(B)}}} \right)$$

               $$ \Rightarrow P({E_1}|{\text{ A}}) = \dfrac{{\dfrac{1}{6} \times \dfrac{3}{4}}}{{\dfrac{1}{6} \times \dfrac{3}{4} + \dfrac{5}{6} \times \dfrac{1}{4}}}$$

               $$ \Rightarrow P({E_1}|{\text{ A}}) = \dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{8} + \dfrac{5}{{24}}}}$$

               $$ \Rightarrow P({E_1}|{\text{ A}}) = \dfrac{{\dfrac{1}{8}}}{{\dfrac{{3 + 5}}{{24}}}}$$

               $$ \Rightarrow P({E_1}|{\text{ A}}) = \dfrac{{1 \times 24}}{{8 \times 8}}$$

               $$ \Rightarrow P({E_1}|{\text{ A}}) = \dfrac{3}{8}$$

               $${\text{The probability that it is actually a six is}}$$ $$ \dfrac{3}{8}.$$

$${\textbf{Hence. option A}}{\textbf{. is the correct answer.}}$$

using Baye's Law,
$$ P(E_A|A)= \dfrac{ _{  }^{ 3 }{  C}_{1} \times \dfrac{ 1 }{ 5 }}{_{  }^{ 3 }{  C}_{1} \times \dfrac{1}{5} +_{  }^{ 3 }{  C}_{1} \times \dfrac{1}{6} + _{  }^{ 3 }{  C}_{1} \times \dfrac{1}{7}}= \dfrac{42}{107}$$

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 envelope are $$TA.$$
We have $$ \displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 2 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 2 } ,\displaystyle P\left( \frac { A }{ { E }_{ 1 } }  \right) =\frac { 2 }{ 8 } ,P\left( \frac { A }{ { E }_{ 2 } }  \right) =\frac { 1 }{ 7 } , $$
Therefore, by Bayes' theorem we have 
$$ \displaystyle \left( \frac { { E }_{ 2 } }{ A }  \right) =\dfrac { P\left( { E }_{ 2 } \right) P\left( \dfrac { A }{ { E }_{ 2 } }  \right)  }{ P\left( { E }_{ 1 } \right) P\left( \dfrac { A }{ { E }_{ 1 } }  \right) +P\left( { E }_{ 2 } \right) P\left( \dfrac { A }{ { E }_{ 1 } }  \right)  } =\frac { 4 }{ 11 }  $$ 

$$P\left( A \right) =0.5,P\left( B \right) =0.3,P\left( C \right) =0.2$$

Let $${ E }_{ 1 }$$ denote the event that the letter from $$TATANAGAR$$ and $${ E }_{ 2 }$$ the event that the letter come from $$CALCUTTA$$.
Let $$A$$ denote the vent that the two consecutive alphabets visible on the envelop are $$TA$$
We have $$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 2 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 2 } ,P\left( \frac { A }{ { E }_{ 1 } }  \right) =\frac { 2 }{ 8 } ,P\left( \frac { A }{ { E }_{ 2 } }  \right) =\frac { 1 }{ 7 } $$
Therefore by Bayes's theorem we have
$$\displaystyle P\left( \frac { { E }_{ 1 } }{ A }  \right) =\frac { P\left( { E }_{ 1 } \right) .P\left( \frac { A }{ { E }_{ 1 } }  \right)  }{ P\left( { E }_{ 1 } \right) .P\left( \frac { A }{ { E }_{ 1 } }  \right) +P\left( { E }_{ 2 } \right) .P\left( \frac { A }{ { E }_{ 2 } }  \right)  } =\frac { 7 }{ 11 } $$

Let us define the following events.
$${ E }_{ 1 }:$$ Disease $$X$$ is diagnosed correctly by doctor $$A$$.
$${ E }_{ 2 }:$$ Disease $$X$$ is not diagnosed correctly by doctor $$A$$.
$$B:$$ A patient (of doctor $$A$$) who has disease $$X$$ dies.
Then, we are given, $$P({ E }_{ 1 })=0.6$$,$$P({ E }_{ 2 })=1-P({ E }_{ 1 })=1-0.6=0.4$$
and $$\displaystyle P\left( \frac { B }{ { E }_{ 1 } }  \right) =0.4$$, $$\displaystyle P\left( \frac { B }{ { E }_{ 2 } }  \right) =0.7$$
By Bay's Theorem
$$\displaystyle P\left( \frac { { E }_{ 1 } }{ B }  \right) =\frac { P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } }  \right)  }{ P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } }  \right) +P\left( { E }_{ 2 } \right) P\left( \frac { B }{ { E }_{ 2 } }  \right)  } $$
$$\displaystyle=P\left( \frac { { E }_{ 1 } }{ B }  \right) =\frac { 0.6\times 0.4 }{ 0.6\times 0.4+0.4\times 0.7 } =\frac { 0.24 }{ 0.24+0.28 } =\frac { 0.24 }{ 0.52 } =\frac { 6 }{ 13 } $$