Probability Test - 27

Let E be the event that either A or B fails.Also let $${ A }^{ F }$$ and $${ B }^{ F }$$ be the event of failure of A and B respectively.
Then $$P\left( E \right) =P\left( { A }^{ F }\cup { B }^{ F } \right) =P\left( { A }^{ F } \right) +P\left( { B }^{ F } \right) -P\left( { A }^{ F }\cap { B }^{ F } \right) \\ \Rightarrow P\left( E \right) =\dfrac { 1 }{ 5 } +\dfrac { 3 }{ 10 } -\left( \dfrac { 1 }{ 5 } \times \dfrac { 3 }{ 10 }  \right) $$
since the failing of A and B are independent.
$$\Rightarrow P\left( E \right) =\dfrac { 10+15-3 }{ 50 } =\dfrac { 22 }{ 50 } =\dfrac { 11 }{ 25 }  $$

$$P$$ (min $$4$$ or max $$8$$) = P(min $$4$$) + P(max $$8$$) - P(min $$4$$ and max $$8$$)
P(min $$4$$)  $$= 6C_2/10C_3$$
P(max 8) $$ = 7C_2/10C_3$$
P(min $$4$$ and max $$8$$)$$ = 3C_1/10C_3$$

$$P = 15/120 + 21/120 - 3/120 = 11/40$$

$$\because P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =P\left( A \right) +P\left( B \right) -P\left( A \right) P\left( B \right) $$      since A and B are independent
putting the values
$$\Rightarrow \dfrac { 7 }{ 10 } =\dfrac { 1 }{ 5 } +P\left( B \right) -\dfrac { P\left( B \right)  }{ 5 } \\ \Rightarrow \dfrac { 4P\left( B \right)  }{ 5 } =\dfrac { 7 }{ 10 } -\dfrac { 1 }{ 5 } =\dfrac { 1 }{ 2 } \\ \Rightarrow P\left( B \right) =\dfrac { 5 }{ 8 } \\ \Rightarrow P\left( { B }^{ c } \right) =1-P\left( B \right) =1-\dfrac { 5 }{ 8 } =\dfrac { 3 }{ 8 } $$

Let $$A(B)$$ denote the event that minimum (maximum) number on the chosen tickets is $$3(7)$$.
We have 
$$P(A)=P($$ choosing $$3$$ and two other numbers from $$4$$ to $$10)$$
$$\displaystyle =\frac { _{  }^{ 7 }{ { C }_{ 2 } } }{ _{  }^{ 10 }{ { C }_{ 2 } } } =\frac { 7\times 6\times 3 }{ 10\times 9\times 8 } =\frac { 7 }{ 40 } $$
$$P(B)=P($$ choosing $$7$$ and choosing two other numbers from $$1$$ to $$6)$$
$$\displaystyle =\frac { _{  }^{ 6 }{ { C }_{ 2 } } }{ _{  }^{ 10 }{ { C }_{ 3 } } } =\frac { 6\times 5\times 3 }{ 10\times 9\times 7 } =\frac { 1 }{ 8 } $$
$$P\left( A\cap B \right) =P($$ choosing $$3$$ and $$7$$ and one other number from $$4$$ to $$6)$$
$$\displaystyle =\frac { 3 }{ _{  }^{ 10 }{ { C }_{ 3 } } } =\frac { 3\times 3\times 3 }{ 10\times 9\times 8 } =\frac { 1 }{ 40 } $$
$$\displaystyle \therefore P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) =\frac { 11 }{ 40 } $$

$$\displaystyle P\left( A \right) =P\left( \frac { A }{ B }  \right) =\frac { P\left( A\cap B \right)  }{ P\left( B \right)  } \Rightarrow P\left( A\cap B \right) =P\left( A \right) P\left( B \right) $$
Thus, $$A$$ and $$B$$ are independent 
Also, $$\displaystyle P\left( \frac { A' }{ B }  \right) =\frac { P\left( A'\cap B \right)  }{ P\left( B \right)  } =\frac { P\left( B \right) -P\left( A\cap B \right)  }{ P\left( B \right)  } $$
$$\displaystyle =1-P\left( \frac { A }{ B }  \right) =1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } $$

$$P\left( A \right) =1-P\left( A' \right) =1-0.3=0.7,P\left( B \right) =0.4$$
We know that 
$$P\left( A \right) =P\left( A\cap B \right) +P\left( A\cap B' \right) \Rightarrow 0.7=P\left( A\cap B \right) =0.5\\ \Rightarrow P\left( A\cap B \right) =0.7-0.5=0.2\\ P\left( A\cup B' \right) =P\left( A \right) +P\left( B' \right) -P\left( A\cap B' \right) =0.7+\left( 1-0.4 \right) -0.5=0.8$$
Next $$\displaystyle P\left( \frac { B }{ \left( A\cup B' \right)  }  \right) =\frac { P\left[ B\cap \left( A\cap B' \right)  \right]  }{ P\left( A\cap B' \right)  } $$
From set theory $$B\cap \left( A\cap B' \right) =\left( B\cap A \right) \cup \left( B\cap B' \right) =\left( B\cap A \right) \cup \phi =B\cap A$$
$$\therefore$$ Required probability $$\displaystyle =\frac { P\left( A\cap B \right)  }{ P\left( A\cup B' \right)  } =\frac { 0.2 }{ 0.8 } =\frac { 1 }{ 4 } $$

$$P($$ at least one event $${ E }_{ 1 },{ E }_{ 2 },{ E }_{ 3 }$$ occurs $$)$$
$$=P\left( { E }_{ 1 }\cup { E }_{ 2 }\cup { E }_{ 3 } \right) \\ =P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) +P\left( { E }_{ 3 } \right) -P\left( { E }_{ 1 }\cap { E }_{ 2 } \right) -P\left( { E }_{ 2 }\cap { E }_{ 3 } \right) \\ -P\left( { E }_{ 3 }\cap { E }_{ 1 } \right) +P\left( { E }_{ 1 }\cap { E }_{ 2 }\cap { E }_{ 3 } \right) \\ =P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) +P\left( { E }_{ 3 } \right) -P\left( { E }_{ 1 } \right) P\left( { E }_{ 2 } \right) -P\left( { E }_{ 2 } \right) P\left( { E }_{ 3 } \right) \\ -P\left( { E }_{ 3 } \right) P\left( { E }_{ 1 } \right) +P\left( { E }_{ 1 } \right) P\left( { E }_{ 2 } \right) P\left( { E }_{ 3 } \right) \\ =0.2+0.4+0.6-\left( 0.2 \right) \left( 0.4 \right) -\left( 0.4 \right) \left( 0.6 \right) -\left( 0.6 \right) \left( 0.2 \right) \\ +\left( 0.2 \right) \left( 0.4 \right) \left( 0.6 \right) =0.808$$

Since $$4$$ cards can be drawn at a time from a pack of $$52$$ cards is $$^{ 52 }{ { C }_{ 4 } }$$ ways,
total number of elementary events $$=^{ 52 }{ { C }_{ 4 } }$$
Consider the following events:
$$A:$$ Getting all spade cards;
$$B:$$ Getting all club cards;
$$C:$$ Getting all diamond cards;
$$D:$$ Getting all hearts cards.
Then $$A,B,C$$ and $$D$$ are mutually exclusive events such that
$$\displaystyle P\left( A \right) =\frac { _{  }^{ 13 }{ { C }_{ 4 } } }{ ^{ 52 }{ { C }_{ 4 } } } ,P\left( B \right) =\frac { _{  }^{ 13 }{ { C }_{ 4 } } }{ ^{ 52 }{ { C }_{ 4 } } } ,P\left( C \right) =\frac { _{  }^{ 13 }{ { C }_{ 4 } } }{ ^{ 52 }{ { C }_{ 4 } } } $$ and $$\displaystyle P\left( D \right) =\frac { _{  }^{ 13 }{ { C }_{ 4 } } }{ ^{ 52 }{ { C }_{ 4 } } } $$
Now required probability
$$=P\left( A\cup B\cup C\cup D \right) $$
$$=P\left( A \right) +P\left( B \right) +P\left( C \right) +P\left( D \right) $$
$$\displaystyle =4\left( \frac { _{  }^{ 13 }{ { C }_{ 4 } } }{ ^{ 52 }{ { C }_{ 4 } } }  \right) =\frac { 44 }{ 4165 } $$

Here given $$P(A\cap B) = \cfrac{1}{6}, P(A\cup B) = \cfrac{2}{3}, P(A)=2P(B)$$
Thus using $$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow \cfrac{2}{3}=P(A)+\cfrac{1}{2}P(A)-\cfrac{1}{6}\Rightarrow P(A) = \cfrac{5}{9}$$