Tag: baye's theorem

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

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