Probability that the ball drawn is red and from ! =$P(R/A)$

Total number of balls = 12
Number of white balls = x
P (white ball) = $= \displaystyle \frac{x}{12}$

$Total\quad number\quad of\quad cards=52\ Number\quad of\quad lost\quad cards=1\ 13\quad cards\quad are\quad surely\quad red\quad therfore,\quad from\quad the\quad remaining\quad 39\quad cards\quad 26\quad are\quad black\quad and\quad 13\quad are\quad red\ So\quad probability\quad of\quad lost\quad card\quad being\quad black\quad =\frac { \left( 26\ 1 \right)  }{ \left( 39\ 1 \right)  } =\frac { 26 }{ 39 } =\frac { 2 }{ 3 } $

Let $X$ be the probability of choosing bag $A$,$Y$ be the probability of choosing bag $B$

There are three odd envelops and three even envelops 

$\begin{array}{l}\left. \begin{array}{l}{\rm{combinations}} = \left( {4,1} \right)\left( {4,2} \right)\\left( {4,3} \right)\left( {4,4} \right)\end{array} \right} \le 7\\left. {\left( {4,5} \right)\left( {4,6} \right)} \right} > 7\{\rm{P}}\left( { > 7} \right) = \frac{1}{3}\end{array}$

\begin{array}{l} P\left( { A/{ E_{ 1 } } } \right) =\dfrac { 5 }{ { 100 } }  \\ P\left( { A/{ E_{ 2 } } } \right) =\dfrac { { 10 } }{ { 100 } }  \\ P\left( { { E_{ 1 } } } \right) =\dfrac { { 60 } }{ { 100 } } ,P\left( { { E_{ 2 } } } \right) =\dfrac { { 40 } }{ { 100 } }  \\ P\left( { A|{ E_{ 1 } } } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right)  } }{ { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { A|{ E_{ 2 } } } \right)  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } }  } }{ { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } +\dfrac { { 40 } }{ { 100 } } \times \dfrac { { 10 } }{ { 100 } }  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 60\times 5 } }{ { 60\times 5+40\times 10 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 300 } }{ { 300+400 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { 3 }{ 7 }  \end{array}

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=$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$

Required probability$\displaystyle =\dfrac {\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {1}{3}+\dfrac {2}{3}.\dfrac {2}{3}\right )}{\dfrac {1}{3}\left (\dfrac {1}{3}.\dfrac {2}{3}.\dfrac {2}{3}\right )+\dfrac {1}{3}\left (\dfrac {2}{3}.\dfrac {3}{6}+\dfrac {1}{3}.\dfrac {3}{6}\right )+\dfrac {1}{3}\left (\dfrac {3}{6}.\dfrac {2}{3}+\dfrac {3}{6}.\dfrac {1}{3}\right )}$

$\displaystyle =\dfrac {\dfrac {5}{9}}{\dfrac {5}{9}+\dfrac {9}{18}+\dfrac {9}{18}}\\ =\dfrac {5}{14}$

$E _1$ : only A hits the target
$E _2$ : only B hits the target
$E$ : exactly one hits the target.
$\therefore \displaystyle P(E _1 / E) = \frac{P(E _1). P (E / E _1)}{P (E _1). P (E/ E _1) + P (E _2). P (E/ E _2)}$
$=

\displaystyle \frac{\displaystyle \frac{9}{10} \times \frac{1}{15}}{

\displaystyle \frac{9}{10} \times \frac{1}{15} + \frac{14}{15} \times

\frac{1}{10}}\ = \dfrac{9}{23}$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

Possible arrangements are BWBW....... or WBWB......

Let $A$ denote the event that the target is hit when $x$ shells are fired at point $I$.
Let ${ E } _{ 1 }$ and ${ E } _{ 2 }$ denote the events hitting $I$ and $II$, respectively
$\displaystyle \therefore P\left( { E } _{ 1 } \right) =\frac { 8 }{ 9 } ,P\left( { E } _{ 2 } \right) =\frac { 1 }{ 9 } $
Now $\displaystyle P\left( \frac { A }{ { E } _{ 1 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x }$ and $\displaystyle P\left( \frac { A }{ { E } _{ 2 } }  \right) =1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }$
Hence $\displaystyle P\left( A \right) =\frac { 8 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ x } \right] +\frac { 1 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x } \right] $
$\displaystyle \therefore \frac { dP\left( A \right)  }{ dx } =\frac { 8 }{ 9 } \left[ { \left( \frac { 1 }{ 2 }  \right)  }^{ x }\log { 2 }  \right] +\frac { 1 }{ 9 } \left[ -{ \left( \frac { 1 }{ 2 }  \right)  }^{ 21-x }\log { 2 }  \right] $
For maximum probability $\displaystyle \frac { dP\left( A \right)  }{ dx } =0$
$\therefore x=12$   $\left[ \because { 2 }^{ 3-x }={ 2 }^{ x-21 }\Rightarrow 3-x=x-21 \right] $
Since $\displaystyle \frac { { d }^{ 2 }P\left( A \right)  }{ dx^{ 2 } } <0$ for $x=12$
$\therefore P\left( A \right) $ is maximum for $x=12$

We define the following events
${ A } _{ 1 }:$ He knows the answer
${ A } _{ 2 }:$ He does not know the answer
$E:$ He gets the correct answer
Thus $\displaystyle P\left( { A } _{ 1 } \right) =\frac { 9 }{ 10 } ,P\left( { A } _{ 2 } \right) =1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } ,P\left( \frac { E }{ { A } _{ 1 } }  \right) =1,P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 4 } $
$\therefore$ required probability $\displaystyle =P\left( \frac { { A } _{ 2 } }{ E }  \right) =\frac { P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right)  }{ P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right)  } $
$\displaystyle =\frac { \dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  }{ \dfrac { 9 }{ 10 } .1+\dfrac { 1 }{ 10 } .\dfrac { 1 }{ 4 }  } =\frac { 1 }{ 36+1 } =\frac { 1 }{ 37 } $

Let $E$ be the event that employee received the letter and $A$ that employer received the reply, then
$\displaystyle P\left ( E \right )= \frac{n-1}{n}$ and $\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$

Let  $\displaystyle E _{1}$ be the event that a subject is selected from first group.
$\displaystyle E _{2}$ the event that a subject is selected from the second group.
$E$ be the event that an engineering subject is selected.
Now the probability that die shows $3$ or $5$  is
$\displaystyle P(E _1)=\frac{2}{6}=\frac{1}{3}$

Let $E _1(0\leq i\leq 2)$ denote the event that urn contains $i$ white and $(2-i)$ black balls.
Let $A$ denote the event that a white ball is drawn from the urn.
We have $P(E _i)=1/3$ for $i=0, 1, 2$. and $P(A|E _1)=1/3, P(A|E _2)=2/3, P(A|E _3)=1$.
By the total probability rule,
$P(A)=P(E _1)P(A|E _1)+P(E _2)P(A|E _2)+P(E _3)P(A|E _3)$
$\displaystyle =\frac {1}{3}\left [\frac {1}{3}+\frac {2}{3}+1\right ]=\frac {2}{3}$

Let $E$ be the event that the man reports that six occurs whle throwing the die and let $S$ be the event that six occurs. Then 
$P(S)=$ Probability that six occurs $ \displaystyle =\frac { 1 }{ 6 }   $
$P\left( { S }^{ 1 } \right) =$ probability that six does not occur $ \displaystyle =1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } $
$ \displaystyle P\left( \frac { F }{ S }  \right) = $ probability that the man reports that six occurs when six has actually occurred 
$=$ probability that the man reports the truth $ \displaystyle =\frac { 3 }{ 4 }  $ 
$ \displaystyle P\left( \frac { E }{ { S }^{ 1 } }  \right) =$ probability that the man report that six occur when six has not actually occurred. 
$=$ probability that the man does not speak the truth 
$ \displaystyle 1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } . $ 
By Bayes' theorem 
$ \displaystyle P\left( \frac { S }{ E }  \right) =$ probability that the man 
reports that six occurs when six has actually occured 
$ \displaystyle =\frac { P\left( S \right) P\left( \frac { F }{ S }  \right)  }{ P\left( S \right) \times P\left( \frac { F }{ S }  \right) +P\left( { S }^{ 1 } \right) \times P\left( \frac { E }{ { S }^{ 1 } }  \right)  }$ 
$ \displaystyle =\frac { \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 }  }{ \dfrac { 1 }{ 6 } \times \dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } \times \dfrac { 1 }{ 4 }  } =\frac { 1 }{ 8 } \times \frac { 24 }{ 8 } =\frac { 3 }{ 8 }  $

Let $E _i$ denote the event that the bag contains $i$ black and ($10-i$) white balls $(i=0, 1, 2, ...., 10)$. Let $A$ denote the event that the three balls drawn at random from the bag are black. We have
$P(E _i)=\dfrac {1}{11} (i=0, 1, 2, ...., 10)$
$P(A|E _i)=0$ for $i=0, 1, 2$
and $P(A|E _i)=\dfrac {^iC _3}{^{10}C _3}$ for $i\geq 3$
Now, by the total probability rule
$\displaystyle P(A)=\sum _{i=0}^{10}P(E _i)P(A|E _i)$
$=\frac {1}{11}\times \frac {1}{^{10}C _3}[^3C _3+^4C _4+....+^{10}C _3]$
But $^3C _3+^4C _3+^5C _3+....+^{10}C _3$
$=^4C _4+^4C _3+^5C _3+...+^{10}C _3$
$=^5C _4+^5C _3+^6C _3+....+^{10}C _3$
$=^6C _4+^6C _3+....+^{10}C _3=....=^{11}C _4$
Thus, $P(A)=\dfrac {^{11}C _4}{11\times ^{10}C _3}=\dfrac {1}{4}$
By the Bayes' rule
$P(E _9|A)=\dfrac {P(E _9)P(A|E _9)}{P(A)}=\dfrac {\dfrac {1}{11}\dfrac {(^9C _3)}{^{10}C _3}}{\dfrac {1}{4}}=\dfrac {14}{55}$.

Let $E _1, E _2$ and $A$ denote the following events:
$E _1$: coin selected is fair
$E _2$: coin selected is biased
$A: $ the first toss results in a head and the second toss results in a tail.
$\displaystyle P(E _1)=\frac {m}{N}, P(E _2)=\frac {N-m}{N}$,
$\displaystyle P(A|E _1)=\frac {1}{2}\times \frac {1}{2}\times \frac {1}{4}, P(A|E _2)=\frac {2}{3}\times \frac {1}{3}=\frac {2}{9}$.
By Bayes' rule
$\displaystyle P(E _1|A)=\frac {P(E _1)P(A|E _1)}{P(E _1)P(A|E _1)+P(E _2)P(A|E _2)}=\frac {9m}{8N+m}$.

$P(A)=\dfrac {m-1}{m}, P(A')=\dfrac {1}{m}$
$P(B|A)=\dfrac {m-1}{m}, P(B|A')=0$
Now $P(B|A)=\dfrac {m-1}{m}\Rightarrow \dfrac {P(A\cap B)}{P(A)}=\dfrac {m-1}{m}$
$\Rightarrow P(A\cap B)=\dfrac {(m-1)^2}{m^2}$
Also $P(B)=P(A) P(B|A)+P(A') P(B|A')$
$=\left (\dfrac {m-1}{m}\right )\left (\dfrac {m-1}{m}\right )+\left (\dfrac {1}{m}\right )(0)=\left (\dfrac {m-1}{m}\right )^2$
$\Rightarrow P(B')=1-P(B)=1-\dfrac {(m-1)^2}{m^2}=\dfrac {2m-1}{m^2}$
$\therefore P(A|B')=\dfrac {P(A\cap B)}{P(B')}=\dfrac {P(A)-P(A\cap B)}{P(B')}$
$=\dfrac {m-1}{2m-1}$
$P(A\cup B)=\dfrac {m-1}{m}+\left (\dfrac {m-1}{m}\right )^2-\left (\dfrac {m-1}{m}\right )^2$.

$P(B)=P(A)P(B|A)+P(A')P(B|A')$
$=P(A)P(B|A)+P(A')[1-P(B'|A')]$
$=\alpha(1-\alpha)+(1-\alpha)[1-(1-\alpha)]=2\alpha (1-\alpha)$

Let ${ A } _{ 1 }$ denote the event that a coin having heads on both sides is chosen, and ${ A } _{ 2 }$ denote the vent that a fiar coin is chosen.
Let $E$ denote the vent that head occurs, Then
$\displaystyle P\left( { A } _{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A } _{ 2 } \right) =\frac { n+1 }{ 2n+1 } $
Probability of occurrence of event $E$, if unfair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 1 } }  \right) =1$
Probability of occurrence of event $E$, if fair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 2 } $
$\because P\left( E \right) =P\left( { A } _{ 1 }\cap E \right) +P\left( { A } _{ 2 }\cap E \right) $
$\displaystyle \therefore P\left( E \right) =P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right) $
$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right)  } \ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$