Probability Test 9

Here $$\displaystyle P\left( A \right) =\frac { 25 }{ 100 } ,P\left( B \right) =\frac { 35 }{ 100 } ,P\left( C \right) =\frac { 40 }{ 100 } $$

Here we have to find $$\displaystyle P\left ( A_{1}/B \right ).$$
By Baye's theorem
$$\displaystyle P\left ( A_{1}/B \right )= \frac{P\left ( A_{1} \right )P\left ( B/A_{1} \right )}{P\left ( A_{1} \right )P\left ( B/A_{1} \right )+P\left ( A_{2} \right )P\left ( B/A_{2} \right )+P\left ( A_{3} \right )P\left ( B/A_{3} \right )}$$
$$\displaystyle = \dfrac{\dfrac{1}{3}.\dfrac{2}{5}}{\dfrac{3}{5}},$$
$$\displaystyle = \frac{2}{9}.$$

Given event $$A$$ and $$B$$ are mutually exclusive 
$$\therefore A\cap B = \phi\Rightarrow n\cap B = 0\Rightarrow P\left(A\cap B\right) = 0$$

$$n(S)\, =\, 4, n(A)\, =\, 3, n(B)\, =\, 1, n(C)\, =\, 2, A \cap B\, =\, \phi$$ 
$$\therefore\, A$$ and $$ B\,$$ are mutually exclusive events.

Let $$W$$ denote the event that $$A$$ draws a white ball and $$T$$ the event that $$A$$ speak truth.
In the usual notations, we are given that 
$$\displaystyle P\left( W \right) =\frac { 1 }{ 9 } ,P\left( \frac { T }{ w }  \right) =\frac { 5 }{ 6 } $$
so that $$\displaystyle P\left( \overline { W }  \right) =1-\frac { 1 }{ 9 } =\frac { 8 }{ 9 } ,P\left( \frac { T }{ \overline { W }  }  \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 6 } $$.
Using Baye's theorem required probability is given by 
$$\displaystyle P\left( \frac { W }{ T }  \right) =\frac { P\left( W\cap T \right)  }{ P\left( T \right)  } =\frac { P\left( W \right) P\left( \frac { T }{ w }  \right)  }{ P\left( W \right) P\left( \frac { T }{ w }  \right) +P\left( \overline { W }  \right) P\left( \frac { T }{ \overline { W }  }  \right)  } $$
$$\displaystyle =\frac { \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 }  }{ \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 } +\dfrac { 8 }{ 9 } \times \dfrac { 1 }{ 6 }  } =\frac { 5 }{ 13 } $$

By the Bayes' rule
$$\displaystyle P(E_1|A)=\frac {P(E_1)P(A|E_1)}{P(A)}=\frac {2}{n(n+1)}$$

Let us define the events:
$$W:$$ Players wins on Monday
$$F:$$ Track is fast
$$S:$$ Track is slow
We have given: $$\displaystyle P\left( F \right) =0.7,P\left( S \right) =0.3,P\left( \frac { W }{ F }  \right) =0.3,P\left( \frac { W }{ S }  \right) =0.4$$
$$\therefore P($$ player $$X$$ will win on Monday $$)$$
$$\displaystyle =P\left( W \right) =P\left( W\cap F \right) +P\left( W\cap S \right) =P\left( F \right) \times P\left( \frac { W }{ F }  \right) +P\left( S \right) \times P\left( \frac { W }{ S }  \right) \\ =0.7\times 0.3+0.3\times 0.4=0.33$$

Let $$A$$ be the event that a really able candidate passes the test 
and let $$B$$ be the event that any candidate passes this test.
Then we have
$$\displaystyle P\left( \frac { B }{ A }  \right) =0.8,P\left( \frac { B }{ A' }  \right) =0.25,P\left( A \right) =0.4,P\left( A' \right) =1-0.4=0.6$$
By Baye's formula
$$\displaystyle P\left( \frac { A }{ B }  \right) =\frac { P\left( A \right) P\left( \frac { B }{ A }  \right)  }{ P\left( A \right) P\left( \frac { B }{ A }  \right) +P\left( A' \right) P\left( \frac { B }{ A' }  \right)  } =\frac { 0.32 }{ 0.32+0.15 } =\frac { 32 }{ 47 } =68$$%

Here, the sample space is $$S=\{1,2,3,4,5,6\}$$
$$\Rightarrow n(S) = 6$$
$$A$$ : Getting a prime number $$\Rightarrow A=\{2,3,5\}$$
$$\therefore n(A) = 3$$
$$B$$ : Getting a number divisible by $$3 \Rightarrow B=\{3,6\}$$
$$\therefore n(B) = 2$$
$$C$$ : Getting a perfect square $$\Rightarrow C=\{1,4\}$$
$$\therefore n(C) = 2$$

Let $$E_1$$ denote the event that an ace occurs and $$E_2$$ the event that it does not occur. Let $$A$$ denote the event that the person reports that it is an ace.