Problems

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Solution:

1. \(\mathbb{P}(\text{head}) = \mathbb{P}(\text{head}|1)\mathbb{P}(\text{coin 1}) + \mathbb{P}(\text{head}|2)\mathbb{P}(\text{coin 2})+\mathbb{P}(\text{head}|3)\mathbb{P}(\text{coin 3}) = \frac13(p_1+p_2+p_3)\)
2. \(N_1 = X_1 + X_2\) where \(X_i \sim Bernoulli(p)\), therefore \(\mathbb{E}(N_1) = 2p\) and \(\textrm{Var}(N_1) = \textrm{Var}(X_1)+ \textrm{Var}(X_2) = p(1-p)+p(1-p) = 2p(1-p)\)
3. Let \(Y_i\) be the indicator for the \(i\)th coin is heads. Then \(\mathbb{E}(Y_i) = p\) and so \(\mathbb{E}(N_2) = 2p\). \begin{align*} && \textrm{Var}(N_2) &= \mathbb{E}(N_2^2) - [\mathbb{E}(N_2)]^2\\ &&&= 2^2 \cdot \left (\frac13 \left (p_1p_2+p_2p_3+p_3p_1 \right) \right) + 1 \cdot \left (\frac13 \left (p_1 (1-p_2) + (1-p_1)p_2 + p_2(1-p_3) +(1-p_2)p_3 + p_3(1-p_1) + (1-p_3)p_1 \right) \right) - [\mathbb{E}(N_2)]^2 \\ &&&= \frac43\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac13 \left ( 2(p_1+p_2+p_3) - 2(p_1p_2+p_2p_3+p_3p_1)\right)-[\mathbb{E}(N_2)]^2 \\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-[\mathbb{E}(N_2)]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-\left[\frac23(p_1+p_2+p_3)\right]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p)\\ \end{align*}
4. \(\,\) \begin{align*} && \textrm{Var}(N_1) - \textrm{Var}(N_2) &= 2p(1-p) - \left (\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p) \right) \\ &&&= 2p^2-\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) \\ &&&= \frac23 \left ( \frac13(p_1+p_2+p_3)^2 -\left (p_1p_2+p_2p_3+p_3p_1 \right)\right)\\ &&&= \frac29 \left (p_1^2+p_2^2+p_3^2 -(p_1p_2+p_2p_3+p_3p_1) \right)\\ &&&= \frac19 \left ((p_1-p_2)^2+(p_2-p_3)^2+(p_3-p_1)^2 \right) &\geq 0 \end{align*} with equality iff \(p_1 = p_2 = p_3\)

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Solution:

1. The probability the first is different to the second is \(\frac68\), the probability the third is different to both of the first two is \(\frac37\) therefore the probability is \(\frac{6}{8} \cdot \frac37 = \frac9{28}\) Whatever pills we are left with on the last day is essentially the same random choice as we make on the first day, therefore \(\frac9{28}\)
2. The probability the first is different to the second is \(\frac{2n}{3n-1}\), the probability the third is different to both of the first two is \(\frac{n}{3n-2}\) therefore the probability is \(\frac{2n^2}{(3n-1)(3n-2)}\). [We can also view this as \(\frac{(3n) \cdot (2n) \cdot n}{(3n) \cdot (3n-1) \cdot (3n-2)}\)]
3. Suppose describe the pills as \(B\) and \(R\) and also number them, then we must have a sequence of the form: \[ B_1R_1 \, B_2R_2 \, B_3R_3 \, \cdots \, B_{n}R_n \] However, we can also rearrange the order of the \(B\) and \(R\) pills in \(n!\) ways each, and also the order of the pairs in \(2^n\) ways. There are \((2n)!\) orders we could have taken the pills out therefore the probability is \begin{align*} && P &= \frac{2^n (n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}{n}} \\ &&&\approx \frac{2^n \cdot 2n \pi \left ( \frac{n}{e} \right)^{2n}}{\sqrt{2 \cdot 2n \cdot \pi} \left ( \frac{2n}{e} \right)^{2n}} \\ &&&= \frac{2^n \sqrt{n \pi} \cdot n^{2n} \cdot e^{-2n}}{2^{2n} \cdot n^{2n} \cdot e^{-2n}} \\ &&&= 2^{-n} \sqrt{n \pi} \end{align*} There is a nice way to think about this question using conditional probability. Suppose we are drawing out of an infinitely supply of \(R\) and \(B\) pills, then each day there is a \(\frac12\) chance of getting different pills. Therefore over \(n\) days there is a \(2^{-n}\) chance of getting different pills. Conditional on the balanced total we see that \begin{align*} && \mathbb{P}(\text{balanced every day} |\text{balanced total}) &= \frac{\mathbb{P}(\text{balanced every day})}{\mathbb{P}(\text{balanced total})} \end{align*} We have already seen the term that is balanced total is \(\frac{1}{2^{2n}}\binom{2n}{n}\), but we can also approximate the balanced total using a normal approximation. \(B(2n, \tfrac12) \approx N(n, \frac{n}{2})\) and so: \begin{align*} \mathbb{P}(X = n) &\approx \mathbb{P}\left (n-0.5 \leq \sqrt{\tfrac{n}{2}} Z + n \leq n+0.5 \right) \\ &= \mathbb{P}\left (- \frac1{\sqrt{2n}} \leq Z \leq \frac{1}{\sqrt{2n}} \right) \\ &= \int_{- \frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \d x \approx \frac{2}{\sqrt{2n}} \frac{1}{\sqrt{2\pi}} \\ &\approx \frac{1}{\sqrt{n\pi}} \end{align*}

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Solution: There are \(\binom{26}3\binom{23}{1}2\) ways to obtain \(3\) pairs There are \(\binom{26}2 \binom{24}3 \cdot 2^3\) ways to obtain \(2\) pairs There are \(\binom{26}1 \binom{25}5 \cdot 2^5\) ways to obtain \(1\) pairs There are \(\binom{26}7 \cdot 2^7\) ways to obtain \(0\) pairs There are \(\binom{52}{7}\) ways to choose our integers, so \begin{align*} && \mathbb{P}(X = 3) &= \frac{\binom{26}{3} \cdot \binom{23}{1} \cdot 2}{\binom{52}{7}} \\ &&&= \frac{7! \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 2}{3! \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46} \\ &&&= \frac{7 \cdot 6 \cdot 5 \cdot 4 }{51 \cdot 2\cdot 49 \cdot 2\cdot 47 \cdot 2} \\ &&&= \frac{ 5 }{17\cdot 7\cdot 47} = \frac{5}{5593} \\ \\ && \mathbb{P}(X = 2) &= \frac{\binom{26}2 \binom{24}3 \cdot 2^3}{\binom{52}{7}} \\ &&&= \frac{220}{5593} \\ \\ && \mathbb{P}(X = 1) &= \frac{\binom{26}1 \binom{25}5 \cdot 2^5}{\binom{52}{7}} \\ &&&= \frac{1848}{5593} \\ \\ && \mathbb{P}(X = 0) &= \frac{\binom{26}7 \cdot 2^7}{\binom{52}{7}} \\ &&&= \frac{3520}{5593} \\ \\ && \mathbb{E}(X) &= \frac{1848}{5593} + 2 \cdot \frac{220}{5593} + 3 \cdot \frac{5}{5593} \\ &&&= \frac{2303}{5593} = \frac{7}{17} \end{align*} Notice we can find the expected value directly: Let \(X_i\) be the random variable the \(i\)th number is discarded. Notice that \(\mathbb{E}(X) = \mathbb{E}\left (\frac12 \left (X_1 +X_2 +X_3 +X_4 +X_5 +X_6 +X_7\right) \right)\) and also notice that each \(X_i\) has the same distribution (although not independent!). Then \begin{align*} &&\mathbb{E}(X) &= \frac72 \mathbb{E}(X_i) \\ &&&= \frac72 \cdot \left (1 - \frac{50}{51} \cdot \frac{49}{50} \cdots \frac{45}{46} \right) = \frac74 \left ( 1 - \frac{45}{51}\right) \\ &&&= \frac72 \cdot \frac{6}{51} \\ &&&= \frac7{17} \end{align*}

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Solution: Without loss of generality, \(C\) will tell \(B\) about the rumour. If \(B\) tells \(D\) then \(D\) can either tell \(A\) or \(C\) at which point either \(A\) is told or the rumour stops spreading.

Therefore \(\mathbb{P}(\text{Arthur rehears}) = 3/4\) For the chances Chandra hears it twice, still WLOG, assume she tells B: So her chance of hearing it twice is \(\frac12\) The person who hears it 3rd has a \(\frac12\) chance of hearing it twice, but the person who hears if 4th has no chance. Therefore Bertha has a \(\frac14\) chance of hearing it twice.