Let $A_i$ be the event that your friend receives an offer from the $i$th company, i=1,2,3,4. Then, by the union bound:

There are ${n \choose 2}$ possible edges in the graph. Let $E_i$ be the event that the $i$th edge is an isolated edge, then \begin{equation} P(E_i)=p(1-p)^{2(n-2)}, \end{equation} where $p$ in the above equation is the probability that the $i$th edge is present and $(1-p)^{2(n-2)}$ is the probability that no other nodes are connected to this edge. By the union bound, we have

Here, it is convenient to use the following inequality: \begin{equation} 1-x \leq e^{-x}, \quad \text{for all} \quad x \in \mathbb{R}. \end{equation} You can prove it by differentiating $f(x)=e^{-x}+x-1$, and showing that the minimum occurs at $x=0$.

Now, we can write

If $X \sim Exponential (\lambda)$, then $EX=\frac{1}{\lambda}$, using Markov's inequality \begin{equation} P\left(X \geq a\right) \leq \frac{EX}{a}=\frac{1}{\lambda a}. \end{equation} The actual value of $P(X \geq a)$ is $e^{- \lambda a}$, and we always have $ \frac{1}{\lambda a} \geq e^{- \lambda a}$.

1. We have $EX=\frac{1}{\lambda}$ and $Var X=\frac{1}{\lambda^2}$. Using Chebyshev's inequality, we have

   $P\left(|X-EX| \geq b\right)$	$\leq \frac{Var(X)}{b^2}$
   	$= \frac{1}{\lambda ^2 b^2}$.

   
   

If $X \sim Exponential(\lambda)$, then \begin{equation} M_X(s)=\frac{\lambda}{\lambda-s}, \quad \text{for} \quad s<\lambda. \end{equation} Using Chernoff bounds, we have \begin{align} P\left(X \geq a\right) &\leq \min_{s>0} \left[e^{-sa}M_X(s)\right]\\ &= \min_{s>0} \left[e^{-sa} \frac{\lambda}{\lambda-s}\right]. \end{align} If $f(s)=e^{-sa} \frac{\lambda}{\lambda-s}$, to find $\min_{s>0} f(s)$ we write \begin{equation} \frac{d}{ds} f(s)=0. \end{equation} Therefore, \begin{equation} s^{*}=\lambda-\frac{1}{a}. \end{equation} Note since $a>EX=\frac{1}{\lambda}$, then $\lambda -\frac{1}{a}>0$. Thus, \begin{equation} P\left(X \geq a\right) \leq e^{-s^{*}a} \frac{\lambda}{\lambda-s^{*}}= a \lambda e^{1-\lambda a}. \end{equation} The real value of $P\left(X \geq a\right)$ is $e^{-\lambda a}$ and we have $e^{-\lambda a} \leq a\lambda e^{1-\lambda a}$, or equivalently, $a \lambda e \geq 1$, which is true since $a> \frac{1}{\lambda}$.

Using $\rho(X,Y) \leq 1$ and $\rho(X,Y)=\frac{Cov(X,Y)}{\sigma_X\sigma_Y}$, we conclude \begin{equation} \frac{EXY-EXEY}{\sigma_X\sigma_Y} \leq 1. \end{equation} Thus

Note that if you use the Cauchy-Schwarz inequality directly, you obtain:

Using Young's inequality, we conclude that for random variables $U$ and $V$ we have \begin{equation} E|UV| \leq \frac{E|U|^p}{p} + \frac{E|V|^q}{q}. \end{equation} Choose $U=\frac{|X|}{\left(E|X|^p\right)^\frac{1}{p}}$ and $V=\frac{|Y|}{\left(E|Y|^q\right)^\frac{1}{q}}$. We obtain

Since g is convex, we have \begin{equation} g(\alpha x+ (1-\alpha)y) \leq \alpha g(x) + (1-\alpha) g(y), \quad \textrm{for all } \alpha \in [0,1]. \end{equation} Therefore, we have

1. $g(x)$	$=\frac{1}{x+1}$,
   $g^{''}(x)$	$= \frac{2}{(1+x)^3} > 0, \hspace{10pt} \textrm{for } x>0.$

   
   Thus $g$ is convex on $(0,\infty)$

   $E\left[\frac{1}{X+1}\right]$	$\geq \frac{1}{1+EX} \quad (\text{Jensen's inequality})$
   	$=\frac{1}{1+10}$
   	$=\frac{1}{11}.$

   
   
2. If we let $h(x)=e^x , g(x)=\frac{1}{1+x}$ then $h$ is convex and non-decreasing and $g$ is convex thus by problem 8, $e^{\frac{1}{x+1}}$ is a convex function, thus

   $E\left[e^{\frac{1}{1+X}}\right]$	$\geq e^{\frac{1}{1+EX}} \quad (\text{by Jensen's inequality})$
   	$=e^{\frac{1}{11}}.$

   
   
3. If $ g(x) = \ln{\sqrt{x}} = \frac{1}{2} \ln{x}$, then $g^{'}(x)=\frac{1}{2x}$ for $x>0$ and $g^{''}(x)=-\frac{1}{2x^2}$. Thus $g$ is concave on $(0,\infty)$. We conclude

   $E\left[\ln{\sqrt X}\right]$	$=E\left[ \frac{1}{2} \ln X \right]$
   	$\leq \frac{1}{2} \ln{EX} \quad \text{(by Jensen's inequality)}$
   	$= \frac{1}{2} \ln 10.$