From af9d07655a6acc46bb0b83bc91dd8907d1819eaa Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Mon, 7 Sep 2020 23:03:29 -0400 Subject: fixed some typos --- houdre/hw1.tex | 25 ++++++++++++++----------- 1 file changed, 14 insertions(+), 11 deletions(-) diff --git a/houdre/hw1.tex b/houdre/hw1.tex index f442282..4baddda 100644 --- a/houdre/hw1.tex +++ b/houdre/hw1.tex @@ -32,7 +32,7 @@ \def\pfr#1#2{\left(\fr#1#2\right)} \def\form{{ 1 + (\fr23)^n \over 2 }} -For $n$ dice, $\pev_n =\form.$ +We assume that for $n$ dice, $\pev_n =\form.$ Rolling an odd or an even number of dice are complements: they are pairwise disjoint, and their union is $\Omega$, so $$\pod_n + \pev_n = \Pr(\ev\cup\od) = \Pr(\Omega) = 1$$ @@ -87,7 +87,8 @@ This set of events is constructed by taking the power set. For each $E_i\in2^A$, $B_i = E_i\cap(x^c\forall x\in(A \setminus E_i))$ Each of these are in $\F$ because $\forall x,y\in\F\to x\cap y,x\setminus y\in\F$, $A \subset \F$. -$\Pr(\cup A) = \sum_{i=1}^{2^n} \Pr(B_i) \leq \sum_{i=1}^n \Pr(A_i)$. +$\Pr(\cup_i^n A_i) = \Pr(\cup_i^{2^n} B_i) = \sum_{i=1}^{2^n} \Pr(B_i) +\leq \sum_{i=1}^n \Pr(A_i)$. This final statement is true because, for all $B_i$, there exists an $A_j$ that corresponds to a set with $B_i$, and each $A_i$ corresponds to a subset of $B$ which have probabilities which sum to the probability @@ -108,7 +109,7 @@ $$\Omega\setminus A_{n+1} \supseteq A\setminus A_{n+1} \Longrightarrow \Pr(\Omega\setminus A_{n+1}) \geq \Pr(A\setminus A_{n+1}) = \Pr(A)-\Pr(A\cap A_{n+1}).$$ Subtracting the last statement from the assumption, (note that the sign -of the inequality being flips flips the comparator) +of the inequality being flipped flips the comparator) $$\Pr(A) \geq 1 - n + \sum_i^n \Pr(A_i) \Longrightarrow \Pr(A\cap A_{n+1}) \geq 1 - n + \sum_i^n\Pr(A_i) - (1 - \Pr(A_{n+1})) = 1 - (n+1) + \sum_i^{n+1}\Pr(A_i).$$ @@ -207,7 +208,7 @@ $1\over2$. The probability that no one is here, if it has come, is $e^{-1}$, and the prior probability that no one is here has already been stated. $$\Pr(\hbox{9:00 bus came}|\hbox{nobody here}) = {\pfr12 e^{-1}\over -\Pr(\hbox{nobody here}) = {2e^{-1} + e^{-2} + e^{-4}\over 4} +\Pr(\hbox{nobody here})} = {2e^{-1} + e^{-2} + e^{-4}\over 4} = 2{e^{-1}\over 2e^{-1} + e^{-2} + e^{-4}} \approx 0.83.$$ This shows that, if nobody is at the bus stop, there is a better than four to one chance that the 9:00 bus has already come (early). @@ -217,7 +218,7 @@ four to one chance that the 9:00 bus has already come (early). Let $E_k$ represent the event that $k$ heads happen before $s$ tails. $$\Pr(E_r|A={\rm head}) = \Pr(E_{r-1}) + (1-\Pr(E_{r-1}))\Pr(E|A={\rm -tail})$$. +tail}).$$ This is because there are two distinct possibilities after a head is flipped: either $r-1$ more heads are flipped (and $E$ has occurred) or that doesn't happen because some number (may be zero, although @@ -225,19 +226,21 @@ the actual number is irrelevant) of heads happen before a tail, which gives $E$ the same probability as happening as if a tail were flipped first. -$$\Pr(E|A={\rm tail}) = (1-\Pr(\neg E_{s-1}))\Pr(E|A={\rm head}).$$ +$$\Pr(E|A={\rm tail}) = (1-\Pr((\neg E)_{s-1}))\Pr(E|A={\rm head}).$$ This is justified because there are again two possibilities: either the streak of tails continues and $E$ doesn't happen, or a head is flipped. Taking the two equations with substitution gives: -$$\Pr(E|A={\rm tail}) = (1-p^{s-1})(p^{r-1} + (1-p^{r-1}) -\Pr(E|A={\rm tail}) \Longrightarrow -\Pr(E|A={\rm tail})({1\over 1-p^{s-1}} + p^{r-1} - 1)) = 1 - p^{r-1}$$ +$$\Pr(E|A={\rm tail}) = (1-(1-p)^{s-1})(p^{r-1} + (1-p^{r-1}) +\Pr(E|A={\rm tail})$$ +$$\Longrightarrow +\Pr(E|A={\rm tail})({1\over 1-(1-p)^{s-1}} + p^{r-1} - 1)) + = 1 - p^{r-1}$$ $$\Longrightarrow \Pr(E|A={\rm tail}) = {1 - p^{r-1}\over{1\over -1-p^{s-1}} + p^{r-1} - 1}.$$ +1-(1-p)^{s-1}} + p^{r-1} - 1}.$$ \q19 -$$\Pr(\hbox{unmatched socks}) = 2{3n\over(3+n)(2+n)}.$$ +$$\Pr(\hbox{unmatched socks}) = 2{3n\over(3+n)(2+n)}$$ $$= 3\Pr(\hbox{matched red}) = 3{6\over(3+n)(2+n)} \longrightarrow n = 3.$$ -- cgit