Problems

Solution:

1. The probability they pull the key out on the \(k\)th attempt will be \(\left ( \frac{n-1}{n} \right)^{k-1} \frac1n\), so we want: \begin{align*} \E[G_1] &= \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \frac1n \\ &= \frac{1}n \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \\ &= \frac1n \frac{1}{\left (1 - \frac{n-1}{n} \right)^2} \\ &= \frac{1}{n} \frac{n^2}{1^2} = n \end{align*}
2. In version 2, the probability the correct key comes out at the \(k\)th attempt is \(\frac1n\) (assume we take out all the keys, then the correct key is equally likely to appear in all of the space). Therefore \(\E[G_2] = \frac1n (1 + 2 + \cdots + n) = \frac{n+1}{2}\)
3. The probability the key comes out on the correct attempt is: \begin{align*} && \mathbb{P}(G_3 = k) &= \frac{n-1}{n} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \cdots \frac{n+k-3}{n+k-2} \cdot \frac{1}{n+k-1} \\ &&&= \frac{n-1}{(n+k-2)(n+k-1)} \\ \\ &&k \cdot \mathbb{P}(G_3 = k) &= \frac{k(n-1)}{(n+k-2)(n+k-1)} \\ &&&= \frac{(n-1)(2-n)}{n+k-2} + \frac{(n-1)^2}{n+k-1} \\ &&&= \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n-k+2} \\ \Rightarrow && \E[G_3] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(G_3 = k) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n+k-2} \right) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} \right) +\underbrace{\sum_{k=1}^{\infty} \frac{n-1}{n-k+2}}_{\to \infty} \\ \end{align*}

Solution:

1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}

Solution: This can either occur via \(N-2\) heads and \(1\) tail in the first \(N-1\) flips, followed by a tail, or \(N-2\) tails and \(1\) head in the first \(N-1\) flips, followed by another head, ie \begin{align*} \mathbb{P}(N = n) &= \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first tail occurs}}p^{n-2}q^2 + \underbrace{\binom{n-1}{1}}_{\text{ways to choose when the first head occurs}}q^{n-2}p^2 \\ &= (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ \\ \mathbb{E}(N) &= \sum_{n=4}^{\infty} n \cdot \mathbb{P}(N = n) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)p^2q^2(p^{n-4}+q^{n-4}) \\ &= \sum_{n=4}^{\infty} n \cdot (n-1)(p^{n-2}q^2+q^{n-2}p^2) \\ &= q^2\sum_{n=4}^{\infty} n(n-1)p^{n-2}+p^2\sum_{n=4}^{\infty} n(n-1)q^{n-2} \\ &= q^2\left ( \sum_{n=2}^{\infty} n(n-1)p^{n-2} -2 \cdot 1 - 3 \cdot 2 \cdot p\right)+p^2\left ( \sum_{n=2}^{\infty} n(n-1)q^{n-2} - 2-6q\right) \\ &= q^2\left ( 2(1-p)^{-3} -2 - 6 p\right)+p^2\left ( 2(1-q)^{-3} - 2-6q\right) \\ &= q^2\left ( 2q^{-3} -2 - 6 p\right)+p^2\left ( 2p^{-3} - 2-6q\right) \\ &= \frac{2}{q} - 2q^2 - 6pq^2+\frac{2}{p} -2p^2-6p^2q \\ &= \frac{2}{q}+\frac2p - 2(p^2+q^2) - 6pq \\ &= \frac{2}{pq} - 2((p+q)^2-2pq) - 6pq \\ &= \frac{2}{pq} - 2 -2pq \\ &= 2 \left (\frac1{pq} - 1 - pq \right) \end{align*}

Solution: We must have a sequence consisting of \(\underbrace{HTT\cdots TH}_{k-1\text{ heads and }n-k\text{ tails}}\underbrace{H}_{k\text{th head}}\). There are \(\binom{n-1}{k-1}\) ways to chose how to place the \(k-1\) heads in the first \(n-1\) flips, and each sequence has probability \(p^{k-1}(1-p)^{n-k}p\) which gives a probability of \(\displaystyle \binom{n-1}{k-1} p^k (1-p)^{n-k}\). Given that it took an even number of tosses to achieve \(k-1\) heads, this is equivalent to the problem of what is the probability that the first head occurs on an even flip, ie \begin{align*} \mathbb{P}(\text{even flip}) &= \mathbb{P}(2\text{nd flip}) +\mathbb{P}(4\text{th flip}) +\mathbb{P}(6\text{th flip}) + \cdots \\ &= (1-p)p + (1-p)^3p + (1-p)^5p + \cdots \\ &= (1-p)p \left ( \sum_{r=0}^\infty (1-p)^{2r}\right) \\ &= \frac{p(1-p)}{1-(1-p)^2} \\ &= \frac{p(1-p)}{2p-p^2} \\ &= \frac{1-p}{2-p} \end{align*} The ways to achieve \(2\) heads on even tosses are \(EEO\), \(EOE\), \(OEE\). The probability of going from \(O\) to \(E\) is the same as the initial probability of an \(O\) flip, etc. Therefore \begin{align*} \mathbb{P}(EEO) &=\left( \frac{1-p}{2-p} \right)^2 \left ( 1- \frac{1-p}{2-p} \right) \\ &= \left( \frac{1-p}{2-p} \right)^2 \left ( \frac{1}{2-p} \right) \\ \mathbb{P}(EOE) &= \left( \frac{1-p}{2-p} \right) \left ( \frac{1}{2-p} \right)^2 \\ \mathbb{P}(OEE) &= \left ( \frac{1}{2-p} \right)^2 \left( \frac{1-p}{2-p} \right)\\ \mathbb{P}(2 \text{ heads on even tosses}) &= \frac{(1-p)^2 + 2(1-p)}{(2-p)^3} \\ &= \frac{(1-p)(2-p)}{(2-p)^3} \\ &= \frac{1-p}{(2-p)^2} \end{align*}