Tag: probability - iii

Let $\displaystyle { E } _{ i }\left( 0\le i\le 2 \right) $ denotes the event that urn contains $i$ white and $2-i$ black balls.

A: number of ace is drawn in the first n draw
B: an ace appear in the ${ \left( n+1 \right)  }^{ th }$ draw
Hence the probability that exactly n cards are dealt before the first ace appear is equal to $P\left( A\cap B \right) $
$\displaystyle P\left( A \right) =\frac { ^{ 48 }{ { C } _{ n } } }{ ^{ 52 }{ { C } _{ n } } } ,P\left( \frac { B }{ A }  \right) =\frac { 4 }{ 52-n } $
$\displaystyle \therefore P\left( A\cap B \right) =P\left( A \right) .P\left( \frac { B }{ A }  \right) =\frac { ^{ 48 }{ { C } _{ n } } }{ ^{ 52 }{ { C } _{ n } } } .\frac { 4 }{ 52-n } $
$\displaystyle =\frac { 48! }{ \left( 48-n \right) !n! } \times \frac { \left( 52-n \right) !n! }{ 52! } \times \frac { 4 }{ 52-n } $
$\displaystyle =\frac { 4\left( 51-n \right) \left( 50-n \right) \left( 49-n \right)  }{ 52\times 51\times 50\times 49 } $

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

If letter came from Clifton there are $6$ pairs of consecutive letters i.e., $cl, li, if, ft, to$, $ON $ in which $ON$ appears only once.
$\displaystyle \therefore $ the chance that this was the legible couple on the Clifton hypothesis $\displaystyle = \frac{1}{6}$
pairs of consecutive letters in the word London are $lo, on, nd, do$, $ON $ in which $ON$ occurs twice.
$\displaystyle \therefore $ the chance that this was the legible couple on the London hypothesis$=2/5.$
$\displaystyle \therefore $ The a posteriori chances that the letter was from Clifton or London are $\displaystyle \frac{1/6}{\dfrac{1}{6}+\dfrac{2}{5}}$ and $\displaystyle \frac{2/5}{\dfrac{1}{6}+\dfrac{2}{5}}$ respectively.
Thus the reqd.chance $\displaystyle = \frac{12}{17}.$

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. 

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 } $

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
$\displaystyle P\left( \frac { A }{ B }  \right) =\frac { P\left( A \right) P\left( \frac { B }{ A }  \right)  }{ P\left( A \right) P\left( \frac { B }{ A }  \right) +P\left( A' \right) P\left( \frac { B }{ A' }  \right)  } =\frac { 0.32 }{ 0.32+0.15 } =\frac { 32 }{ 47 } =68$%

Given, $A$ is the event which selects the rigged one and $B$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $P(A/E)=\dfrac{1}{4}(1)^3=\dfrac{1}{4}$ ($\dfrac{1}{4}$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $P(B/E)=\dfrac{3}{4}(\dfrac{1}{6})^3=\dfrac{1}{1152}$ (since probability of getting 5 in fair dice case=$\dfrac{1}{5}$)
By Baye's theorem,Probability of selecting the rigged case among both=$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$