Probability Test - 53

$$P(A)=1-P(\overline { A } );P(B)=1-P(\overline { B } )$$
$$P(\overline { AB } )=P(\overline { A } )\times P(\overline { B } )=\left[ 1-P( { A } ) \right] \left[ 1-P( { B } ) \right] $$

Probability of A = Probability of B = Probability of C = $$\dfrac { 1 }{ 3 } $$
Probability of X = $$P(X)=\dfrac { 5 }{ 12 }=P(A)*P(\dfrac { X }{ A } )+P(B)*P(\dfrac { X }{ B } )+P(C)*P(\dfrac { X }{ C } )=\dfrac { 1 }{ 3 } *\dfrac { 3 }{ 8 } +\dfrac { 1 }{ 3 } *\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 3 } *P(\dfrac { X }{ C })$$

It is given that 
$$P\left( A\cap \overline { B }  \right) +P\left( \overline { A } \cap B \right) =p,P\left( B\cap \overline { C }  \right) +P\left( \overline { B } \cap C \right) =p,P\left( A\cap \overline { C }  \right) +P\left( \overline { A } \cap C \right) =p$$
and $$P\left( A\cap B\cap C \right) ={ p }^{ 2 }$$

Since $$1\ge A\cup B\cup C\ge 0.75$$, we have from the diagram that: $$P(B\cup C)=P(A\cup B\cup C)-0.08$$
Hence, $$0.92\ge P(B\cup C)\ge 0.67$$
Now, $$P(B\cap C)=P(B)+P(C)-P(B\cup C)=1.2-P(B\cup C)$$
Hence, $$1.2-0.67\ge P(B\cap C)\ge 1.2-0.92\quad =>\quad 0.53\ge P(B\cap C)\ge 0.28$$

Let $$P(k)$$ = Probability of getting $$k$$ hearts among 5  drawn cards 
$$P(k) = \dfrac{^{13}C_k \times ^{39}C_{5-k}}{^52C_k}$$
i.e choosing k cards out of 13 Heart cards and choosing remaining 5-k cards from others
$$P( \ge 3. hearts) = P(3.hearts)+P(4.hearts)+P(5. hearts)$$

$$\displaystyle= \frac{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}{^{52}C_{5}}$$
.
Now using definition of conditional probability
$$P(k=5| k \ge 3) =  P(k=5) /P(k \ge 3) $$
$$\displaystyle =\frac{^{13}C_{5}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$$

According to the question,
$$P(A) + P(B)-2P(A \cap B) = P(B)+P(C)-2P(B \cap C)= P(A) + P(C) -2P(A \cap C )= p$$.
Adding, we get,  $$2(P(A)+P(B)+P(C)-P(A \cap B)  - P(B \cap C) - P(A \cap C))=3p$$ $$\therefore P(A) + P(B) + P(C)-P(A \cap B) - P(B \cap C) - P(A \cap C)=\displaystyle\frac {3p} {2}$$.

Let the three critics be $$A,B$$ and $$C$$.
The odds that a book will be reviewed favorably by $$A,B$$ and $$C$$ are $$5$$ to $$2,4$$ to $$3$$ and $$3$$ to $$4$$ respectively.
i.e. $$\displaystyle P\left( A \right) =\frac { 5 }{ 7 } ,P\left( B \right) =\frac { 4 }{ 7 } ,P\left( C \right) =\frac { 3 }{ 7 } $$
$$\displaystyle \Rightarrow P\left( \overline { A }  \right) =\frac { 2 }{ 7 } ,P\left( \overline { B }  \right) =\frac { 3 }{ 7 } ,P\left( \overline { C }  \right) =\frac { 4 }{ 7 } $$
To find the probability that a majority will be favorable is the same as the probability that at least two of them are favorable.
$$\therefore$$ the required probability
$$=P\left( A\cap B\cap \overline { C }  \right) +P\left( A\cap \overline { B } \cap C \right) +P\left( \overline { A } \cap B\cap C \right) +P\left( A\cap B\cap C \right) \\ =P\left( A \right) P\left( B \right) P\left( \overline { C }  \right) +P\left( A \right) P\left( \overline { B }  \right) P\left( C \right) +P\left( \overline { A }  \right) P\left( B \right) P\left( C \right) +P\left( A \right) P\left( B \right) P\left( C \right) $$
$$\displaystyle =\frac { 2 }{ 7 } \times \frac { 4 }{ 7 } \times \frac { 4 }{ 7 } +\frac { 5 }{ 7 } \times \frac { 3 }{ 7 } \times \frac { 3 }{ 7 } +\frac { 2 }{ 7 } \times \frac { 4 }{ 7 } \times \frac { 3 }{ 7 } +\frac { 5 }{ 7 } \times \frac { 4 }{ 7 } \times \frac { 3 }{ 7 } =\frac { 209 }{ 343 } $$ 

Let's denote the event of getting a sum of 5 be represented as 5 and that of 7 by 7
hence 5 can be obtained by the combination {1,4 and 2,3} and 7 by {1,6  2,5  3,4} 
thus $$P\left( 5 \right) =\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) \left[ P\left( 1 \right) P\left( 4 \right) +P\left( 2 \right) P\left( 3 \right)  \right] =2\times \left[ \dfrac { 1 }{ 6 } \times \dfrac { 1 }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { 1 }{ 6 }  \right] =\dfrac { 1 }{ 9 } \\ similarly\quad \\ P\left( 7 \right) =\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) \left[ P\left( 1 \right) P\left( 6 \right) +P\left( 2 \right) P\left( 5 \right) +P\left( 3 \right) P\left( 4 \right)  \right] =2\times \left[ \dfrac { 1 }{ 6 } \times \dfrac { 1 }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { 1 }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { 1 }{ 6 }  \right] =\dfrac { 1 }{ 6 } \\ and\quad P\left( \overline { 5 } \cap \overline { 7 }  \right) =1-P\left( 5\cup 7 \right) =1-P\left( 5 \right) -P\left( 7 \right) +0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =1-\dfrac { 1 }{ 9 } -\dfrac { 1 }{ 6 } =\dfrac { 13 }{ 18 } $$
now Let the desired event be represented by E
then $$P\left( E \right) =P\left( 5 \right) +P\left( \overline { 5 } \cap \overline { 7 }  \right) P\left( 5 \right) +P\left( \overline { 5 } \cap \overline { 7 }  \right) P\left( \overline { 5 } \cap \overline { 7 }  \right) P\left( 5 \right) +........................\infty $$
clearly this is the infinite sum of a G.P with first term=$$P\left( 5 \right) $$ and common ratio=$$P\left( \overline { 5 } \cap \overline { 7 }  \right) $$
$$\Rightarrow P\left( E \right) =\dfrac { P\left( 5 \right)  }{ 1-P\left( \overline { 5 } \cap \overline { 7 }  \right)  } =\dfrac { \dfrac { 1 }{ 9 }  }{ 1-\dfrac { 13 }{ 18 }  } =\dfrac { 2 }{ 5 } $$

$$\displaystyle P\left( \alpha  \right) =\frac { 4 }{ 5 } ,P\left( \overline { \alpha  }  \right) =1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } $$

$$\displaystyle P\left( \beta  \right) =\frac { 3 }{ 4 } ,P\left( \overline { \beta  }  \right) =1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } $$

$$\displaystyle P\left( \gamma  \right) =\frac { 5 }{ 6 } ,P\left( \overline { \gamma  }  \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 4 } $$

$$\displaystyle P\left( \delta  \right) =\frac { 2 }{ 3 } ,P\left( \overline { \delta  }  \right) =1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } .$$