Probability Distributions

Given: first $30$ natural numbers

The women can be seated in $^{2n}C _n$ ways. Let A denote the event that pairs occupy the seats (1, 2), (3, 4), ...., (2n-1, 2n)
and B denote the event that pairs occupy the seats (2, 3), (4, 5), (6, 7), ...., (2n-2, 2n-1), (2n-1).
The number of cases favourable to A (B) is $2^n$. (For each man in the pair there are two choices.)
However, there are just two cases common to A and B. One is the case $\left {(M, W), (M, W), ...(M, W)\right }$ of A(B) and $\left {(W, M), (W, M), ..., (W, M)\right }$ of B(A).
Therefore, $P(A\cup B)=P(A)+P(B)-P(A\cup B)$
$\displaystyle =\frac {2^n+2^n-2}{^{2n}C _n}=\frac {2^{n+1}-2}{^{2n}C _n}$.

A discrete variable is one that can only take specific, countable values. Since you cannot have a fraction of a patient, the number of patients is a discrete variable.

This is a conditional probability problem involving arrangements of digits. The constraints on the smallest and largest digits restrict the sample space significantly.

\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}

$E _1$ : 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}$

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 $\displaystyle P\left ( \bar{E} \right )= \frac{1}{n}$