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\title{18.200 Homework 2}
\author{Clinton Reid, R03}
\date{February 18, 2020}

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\maketitle

\section{Problem 2}
\textbf{(a) What is the probability of event A, that any given baby is a boy born on a Monday?}
\[P(\text{Baby is a boy born on a Monday})=\frac{1}{2}\cdot\frac{1}{7}=\frac{1}{14}\]
\textbf{(b) Consider two siblings. Given that at least one of them is a boy, what is the probability that they are both boys?}
\[P(\text{2 boys}|\geq \text{1 boy})=\frac{P(\text{2 boys} \bigwedge \text{1 boy})+P(\text{2 boys} \bigwedge \text{2 boys})}{P(\text{1 boy})+P(\text{2 boys})}\]
\[=\frac{0+\frac{1}{4}}{\frac{1}{2}+\frac{1}{4}}=\frac{1}{3}\]
\\\\Similarly, enumerating the possibilities, given that there is one boy,
\[BG\ GB\ BB\]The probability that both are boys = $\frac{1}{3}$.
\\\\\textbf{(c) Consider two siblings. Given that at least one of them is a boy born on a Monday, what is the probability that the other is also a boy (born on any day of the week)?}
\[P(\text{Other child is a boy}|\text{One of them is a boy born on a Monday})\]
Enumerating the possibilities, you get something like this:
\[B^MG^{1-7}\]\[G^{1-7}B^M\]\[B^MB^{1-7}\]\[B^{1-7}B^M\] This represents 27 total possibilities, because the $B^MB^M$ combination is counted twice in the sample space. And, there are 13 which include two boys once the over-count is corrected for. Therefore, the probability of this is $\frac{13}{27}$.
\section{Problem 3}
\textbf{(a)} $|C_n|=$
\\\\\textbf{(b)} $|D_n|=$
\\\\\textbf{(c)}
\begin{center}
 \begin{tabular}{||c c c||}
 \hline
 k & $C_k$ & $D_k$ \\ [0.5ex]
 \hline\hline
 0 &  & 9 \\
 \hline
 1 & 7 & 78 \\
 \hline
 2 & 545 & 778 \\
 \hline
 3 & 545 & 18744 \\
 \hline
 4 & 545 & 18744 \\
 \hline
 5 & 545 & 18744 \\
 \hline
 6 & 545 & 18744 \\
 \hline
 7 & 545 & 18744 \\
 \hline
 8 & 545 & 18744 \\
 \hline
 9 & 88 & 788\\ [1ex]
 \hline
\end{tabular}
\end{center}
\\\\\textbf{(d)}
\section{Problem 4}
\textbf{(a) What is $Pr(X \geq 1)$, i.e. what is the probability that there are 5 consecutive tosses resulting in THHHT.}
\\\\Since we are looking for the occurrence of THHHT in that particular prder, we can treat it as a single element in the sequence. With this change, we now have 6 positions the THHHT sequence can occupy. \[Pr(X\geq1)=\frac{6*2^5}{2^{10}}=\frac{6}{32}\]
The breakdown of this is:\[Pr(X\geq1)=Pr(X=1)+Pr(X=2)=\frac{6*2^5-1}{2^{10}}+\frac{1}{2^{10}}\]
\\\textbf{(b) What is E[X] and Var(X)?}
\[E[X]=\frac{6*2^5-1}{2^{10}}*1+\frac{1}{2^{10}}*2=0.188\]
\[Var(X)=E[X^2]-E[X]^2=(\frac{6*2^5-1}{2^{10}}*1+\frac{1}{2^{10}}*4)-(0.188)^2=0.155\]
\section{Problem 5}
\textbf{(a) What is Var(X)?}
\[Var(X)=E[X^2]-E[X]^2\]Let $X_i$ be an indicator variable: 1 if $\sigma(i)=i\text{,}  \text{ and 0 otherwise}\ X=\sum_{i=1}^{n} X_i$.
\[E[X^2]=E\Bigg[\Bigg(\sum_{i=1}^{n}X_i\Bigg)^2\Bigg]\]
\[=\Bigg(\sum_{i=1}^{n} E[X_i^2]\Bigg)+\Bigg(\sum_{i=1}^{n} \sum_{j\neq i}E[X_iX_j]\Bigg)\]
\[=1+\Bigg(\sum_{i=1}^{n} \sum_{j\neq i}Pr[X_i=1]E[X_iX_j|X_i=1]\Bigg)\]
\[=1+\Bigg(\sum_{i=1}^{n} \sum_{j\neq i}\frac{1}{n}\frac{1}{n-1}\Bigg)\]
\[=1+1=2\]
\[Var(X)=2-1=1\]
\\\textbf{(b) Use Chebyshev’s inequality to give an upper bound on $P[X\geq t]$ for any given integer $t\geq2$.)} Chebyshev's inequality:
\[P(|X-\mu|\geq c\sigma)\leq \frac{1}{c^2}\]
\[P(X\geq t)=P(|X-1|\geq t-1) \leq \frac{1}{(t-1)^2}\]
\section{\textbf{Simpson's Paradox.}}
\textbf{(a)}$P(\text{Admitted given male})=50\%$\\\\$P(\text{Admitted given female})=33\frac{1}{3}\%$\\\\$P(\text{Admitted to Dept A given male})=70\%$\\\\$P(\text{Admitted to Dept B given male})=10\%$\\\\$P(\text{Admitted to Dept A given female})=75\%$\\\\$P(\text{Admitted to Dept B given female})=12.5\%$
\\\\\textbf{(b)} Although the rates of acceptance in each department were across the board higher for women, a larger number of women were still rejected. This is because a larger number of women applied to the more selective Department B, and consequently a larger number of them were rejected. The rate, in this instance, doesn't give a good picture of the situation, because it does not account for differences in applicant numbers. Sure, a larger proportion of women were accepted in each department, but also, a smaller total number were accepted.
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