Probability Test - 54

$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \\ \because \dfrac { 3 }{ 4 } \le P\left( A\cup B \right) \le 1\\ \Rightarrow \dfrac { 3 }{ 4 } \le P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \le 1\\ \Rightarrow \dfrac { 3 }{ 4 } +P\left( A\cap B \right) \le P\left( A \right) +P\left( B \right) \le 1+P\left( A\cap B \right) \\ \Rightarrow \dfrac { 3 }{ 4 } +\dfrac { 1 }{ 8 } \le P\left( A \right) +P\left( B \right) \le 1+\dfrac { 3 }{ 8 } \\ \Rightarrow \dfrac { 7 }{ 8 } \le P\left( A \right) +P\left( B \right) \le \dfrac { 11 }{ 8 } $$
both the statements are correct but statement $$1$$ is not the correct explanation of statement $$2$$

Let $$x$$ denote the number of tosses required.
Then $$P\left( x=r \right) ={ \left( 1-P \right)  }^{ r-1 }P;\quad r=1,2,3....$$
Let $$E$$ denote the event that the number of tosses required is even. Then,
$$P\left( E \right) =P\left\{ \left( x=2 \right) \cup \left( x=4 \right) \cup \left( x=6 \right) \cup ... \right\} \\ =P\left( x=2 \right) +P\left( x=4 \right) +P\left( x=6 \right) +....\\ =p\left( 1-p \right) +p{ \left( 1-p \right)  }^{ 3 }+p{ \left( 1-p \right)  }^{ 5 }+..$$
$$\displaystyle =\frac { p\left( 1-p \right)  }{ 1-{ \left( 1-p \right)  }^{ 2 } } =\frac { p\left( 1-p \right)  }{ 2p-{ p }^{ 2 } } =\frac { 1-p }{ 2-p } $$
As we are given that $$\displaystyle P\left( E \right) =\frac { 2 }{ 5 } $$, we get
$$\displaystyle 5\left( 1-p \right) =2\left( 2-p \right) \Rightarrow 1-3p=0\Rightarrow p=\frac { 1 }{ 3 } $$

$$P\left( A \right) =\dfrac { 3x+1 }{ 3 } ,P\left( B \right) =\dfrac { 1-x }{ 4 } ,P\left( C \right) =\dfrac { 1-2x }{ 2 } \\ \because 0\le P\le 1\\ \Rightarrow 0\le \dfrac { 3x+1 }{ 3 } \le 1\\ \Rightarrow 0\le 3x+1\le 3\\ \Rightarrow \dfrac { -1 }{ 3 } \le x\le \dfrac { 2 }{ 3 } \\ similarly\quad 0\le \dfrac { 1-x }{ 4 } \le 1\\ \Rightarrow -4\le x-1\le 0\\ \Rightarrow -3\le x\le 1\\ similarly\quad 0\le \dfrac { 1-2x }{ 2 } \le 1\quad \quad \Rightarrow -2\le 2x-1\le 0\quad \\ \Rightarrow \dfrac { -1 }{ 2 } \le x\le \dfrac { 1 }{ 2 } \\ Also\quad 0\le P\left( A\cup B\cup C \right) \le 1\\ \Rightarrow 0\le P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) -P\left( C\cap A \right) +P\left( A\cap B\cap C \right) \le 1\\ \Rightarrow 0\le P\left( A \right) +P\left( B \right) +P\left( C \right) \le 1$$
Since A ,  B and C are mutually exhaustive hence $$P\left( A\cap B \right) =P\left( B\cap C \right) =P\left( C\cap A \right) =P\left( A\cap B\cap C \right) =0$$
$$\Rightarrow 0\le \dfrac { 3x+1 }{ 3 } +\dfrac { 1-x }{ 4 } +\dfrac { 1-2x }{ 2 } \le 1\\ \Rightarrow 0\le \dfrac { 13-3x }{ 12 } \le 1\quad \quad \Rightarrow -12\le 3x-13\le 0\quad \quad \quad \quad \Rightarrow \dfrac { 1 }{ 3 } \le x\le \dfrac { 13 }{ 3 } $$
comparing all the inequalities of x and taking the intersection we get 
$$\dfrac { 1 }{ 3 } \le x\le \dfrac { 1 }{ 2 } \quad or\quad x\in \left[ \dfrac { 1 }{ 3 } ,\dfrac { 1 }{ 2 }  \right] $$

Let $$\displaystyle p_{1}, p_{2}, p_{3}$$ denote the probabilities that first, second and third critics review the book.
Given odds in favour that first critic will review the book is $$5:2$$
 $$\displaystyle p_{1}=\dfrac{5}{5+2}=\dfrac{5}{7}$$
$$\displaystyle  \Rightarrow \bar{p_{1}}=1-\frac{5}{7}=\frac{2}{7}$$

Given odds in favour that second critic will review the book is $$4:3$$
$$p_{2}=\dfrac{4}{4+3}=\dfrac{4}{7}$$
$$\displaystyle  \Rightarrow \bar{p_{2}}=1-\frac{4}{7}=\frac{3}{7}$$

Given odds in favour that third critic will review the book is $$3:4$$
$$p_{3}=\dfrac{3}{3+4}=\dfrac{3}{7}$$
$$\displaystyle  \Rightarrow \bar{p_{3}}=1-\frac{3}{7}=\frac{4}{7}$$

Now the majority will be favourable if any of the two critics is favourable and third is unfavourable or all the three critics are favourable.
Hence the required probability is
$$\displaystyle =p_{1}\:p_{2}\bar{p_{3}}+p_{1}\:\bar{p_{2}}\:p_{3}+\bar{p_{1}}\:p_{2}\:p_{3}+p_{1}\:p_{2}\:p_{3}$$

Let A and B denote the following events:
A:minimum of the chosen number is 3
B: maximum of the chosen number is 7
We have,
$$P(A)=P$$(choosing 3 and two other numbers from 4 to 10)
$$=\cfrac { ^{ 7 }{ { C }_{ 2 } } }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 7\times 6 }{ 2 } \times \cfrac { 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 7 }{ 40 } $$
$$P(B)=P$$(choosing 7 and two other numbers from 1 to 6)
$$=\cfrac { ^{ 6 }{ { C }_{ 2 } } }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 6\times 5 }{ 2 } \times \cfrac { 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 1 }{ 8 } $$
$$P\left( A\cap B \right) =P$$(Choosing 3 and 7 and one other number from 4 to 6)
$$=\cfrac { 3 }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 3\times 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 1 }{ 40 } $$
Now $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \\ =\cfrac { 7 }{ 40 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 40 } =\cfrac { 11 }{ 40 } $$

$${ r }_{ 1 },{ r }_{ 2 },{ r }_{ 3 }\in \left\{ 1,2,3,4,5,6 \right\} $$
$${ r }_{ 1 },{ r }_{ 2 },{ r }_{ 3 }$$ are if the form $$3k,3k+1,3k+2$$
Required probability $$\displaystyle \frac { 3!\times _{  }^{ 2 }{ { C }_{ 1 } }\times _{  }^{ 2 }{ { C }_{ 1 } }\times _{  }^{ 2 }{ { C }_{ 1 } } }{ 6\times 6\times 6 } =\frac { 6\times 8 }{ 216 } =\frac { 2 }{ 9 } $$ 

Let $$\displaystyle A_{th}$$ denote the event that the $$\displaystyle i_{th}$$ letter is placed in the right envelope.
Then the required probability is $$\displaystyle \frac{P\left ( \bar{A}_{1}\cap \bar{A}_{2}\cap .....\cap \bar{A}_{n} \right )}{P\left ( A_{i}\cup A_{2}\cup ......A_{n} \right )}$$.
[By De-Morgan law]
$$\displaystyle =1-P\left ( A_{1}\cup A_{2}\cup....\cup A_{n}  \right )$$
$$\displaystyle =1-\left [ \sum P\left ( A_{i} \right )-\sum p\left ( A_{i}\cap A_{j} \right ) \right ]+\sum P\left ( A_{i}\cap a_{j}\cap A_{k} \right ) i\neq j  i\neq j\neq k $$
$$\displaystyle -\cdots +\left ( -1 
\right )^{n-1}p\left ( A_{i}\cap A_{2}\cap \cdots \cap A_{k} \right )$$
Now $$\displaystyle p\left ( A_{i} \right )=\frac{\left ( n-1 \right )!}{n!}$$ as having placed $$\displaystyle i^{th}$$ letter in the right envelope, the remaining letters can be placed in $$(n-1)!$$ ways.
Similarly, $$\displaystyle P\left ( A_{1}\cap 
A_{2}\cap .....\cap A_{r} \right )=$$ Prob.of $$r$$ particular letters in right envelopes $$\displaystyle =\frac{\left ( n-r \right )!}{n!}$$
$$\displaystyle \therefore 
\sum p\left ( A_{1}\cap A_{2}\cap .....\cap A_{r} \right )$$
$$\displaystyle =^{n}C_{r}.\frac{\left ( n-r \right )!}{n!}=\frac{1}{r!}$$ where $$\displaystyle r=1, 2, 
3, ....n.$$
$$\displaystyle \therefore \sum \left ( \bar{A_{1}}\cap \bar{A}_{2}\cap \cdots \cap\bar{A}_{n}  \right )$$
$$\displaystyle =1-\left \{ \frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\cdots +\left ( -1 \right )^{n}.\frac{1}{n!} \right \}$$
which is equal to first $$n-2$$ terms in the expansion of $$\displaystyle 
e^{-1}$$.

Total number of balls $$=15$$
Total number of ways of selecting $$3$$ balls out of $$15$$ balls is $$^{15}C_3$$.
Number of black balls $$=6$$
Number of non-black balls $$=9$$ (4 white, 5 red)
Probability of no ball drawn  is black $$=\dfrac{^9C_3}{^{15}C_3}=\dfrac{12}{65}$$
Probability of exactly 2 black $$=\dfrac{^6C_2\times ^9C_1}{^{15}C_3}=\dfrac{27}{91}$$
Probability of all of same colour i.e. $$RRR\ or\ WWW\ or\ BBB$$
$$=\displaystyle \frac{^{4}C_{3}+^{5}C_{3}+^{6}C_{3}}{^{15}C_{3}}$$
$$\displaystyle =\frac{4+10+20}{\dfrac{\left ( 15.14.13 \right )}{\left ( 3! \right )}}=\frac{34}{455}$$

Given $$P(A\cup B)=\displaystyle\dfrac{1}{2}$$ and $$P(A\cap B)=\displaystyle\dfrac{2}{5}$$
Fact $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow P(A)+P(B)=P(A\cup B)+P(A\cap B)$$  $$=\displaystyle \dfrac{1}{2}+\dfrac{2}{5}=\cfrac{9}{10}$$
$$\Rightarrow 1-P(A^{c})+1-P(B^{c})=\cfrac{9}{10}$$
$$\Rightarrow P(A^{c})+P(B^{c})= 2-\cfrac{9}{10}=\cfrac{11}{10}$$

We know that $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$