Probability Test - 19

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
$$\displaystyle P\left( \frac { { E }_{ 1 } }{ B }  \right) =\frac { P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } }  \right)  }{ P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } }  \right) +P\left( { E }_{ 2 } \right) P\left( \frac { B }{ { E }_{ 2 } }  \right)  } $$
$$\displaystyle=P\left( \frac { { E }_{ 1 } }{ B }  \right) =\frac { 0.6\times 0.4 }{ 0.6\times 0.4+0.4\times 0.7 } =\frac { 0.24 }{ 0.24+0.28 } =\frac { 0.24 }{ 0.52 } =\frac { 6 }{ 13 } $$

There are $$4$$ cases which start with a Head:
(i) $$HTH$$
(ii)$$ HHT $$
(iii)$$ HHH $$
(iv) $$HTT$$
Probability of each case = $$\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } \times\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 8 } $$
Three of the above four cases have at least two heads.
Hence, required probability = $$\dfrac { 3\times\dfrac { 1 }{ 8 }  }{ 4\times\dfrac { 1 }{ 8 }  } =\dfrac { 3 }{ 4 } $$

$$P(A|B) = P(A \cap B)/P(B)$$
Here $$B = $$getting black ball, $$A =$$ ball is from urn $$A$$
Hence, $$ P(A|B) = \dfrac{1}{2}\times \dfrac{\dfrac{3}{5}}{(1/2\times3/5+1/2 \times 4/7)} = 21/41$$
$$P(\overline{ A } |B)=1-P(A|B)$$
$$\Rightarrow P(\overline { A } |B)=1-P(A|B)$$ $$= 1-21/40= 20/41$$

Since $$A$$ and $$B$$ are independent events, therefore 
$$\displaystyle P\left( A\cap B \right) =P\left( A \right) P\left( B \right) ,P\left( \frac { A }{ B }  \right) =P\left( A \right) $$ and $$\displaystyle P\left( \frac { \overline { A }  }{ B }  \right) =P\left( \overline { A }  \right) .$$
Now,$$P\left( A\cap B \right) =P\left( A \right) P\left( B \right) \neq 0$$
$$\Rightarrow A$$ and $$B$$ are not mutually exclusive 
Since $$A\cap B$$ and $$A\cap \overline { B } $$ are mutually exclusive events such that 
$$\left( A\cap B \right)  \cup \left( A\cap \overline { B }  \right) =A.$$ Therefore,
$$P\left( A\cap B \right) +P\left( A\cap \overline { B }  \right) =P\left( A \right) $$
$$\Rightarrow P\left( A \right)P\left( B \right) +P\left( A\cap \overline { B }  \right) =P\left( A \right) $$
$$\Rightarrow P\left( A\cap \overline { B }  \right) =P\left( A \right) -P\left( A \right) P\left( B \right) =P\left( A \right) \left( 1-P\left( B \right)  \right) =P\left( A \right) P\left( \overline { B }  \right) $$ 
So, $$A$$ and $$\overline { B } $$ are independent events. 
Similarly, $$\overline { A } $$ and $$B$$ are also independent events.
Now, $$\displaystyle P\left( \frac { A }{ B }  \right) =P\left( A \right) $$ and $$\displaystyle P\left( \frac { \overline { A }  }{ B }  \right) =P\left( \overline { A }  \right) $$
$$\displaystyle \Rightarrow P\left( \frac { A }{ B }  \right) +P\left( \frac { \overline { A }  }{ B }  \right) =P\left( A \right) +P\left( \overline { A }  \right) =1.$$

$$P(A) = 50/100 = 1/2$$
$$P(B)= 50/100 = 1/2$$
$$P(C)= ( 1/2)(1/2 ) +( 1/2)(1/2) = 1/2$$
$$P(A\cap B) = P(A).P(B) = 1/4$$
$$P(B\cap C)= P(B).P(C)=1/4$$
$$P(A\cap  C)= P(A).P(C)=1/4$$
$$P(A \cap B \cap C) = 0$$

If $$1200$$ are above $$30,$$ there are $$5000-1200=3800$$ which are under $$30$$ 

A derangement is a permutation of objects that leave no object in its original position.

$$^{18}C_3-^{5}C_3=806$$