5 problems found
A game in a casino is played with a fair coin and an unbiased cubical die whose faces are labelled \(1, 1, 1, 2, 2\) and \(3.\) In each round of the game, the die is rolled once and the coin is tossed once. The outcome of the round is a random variable \(X\). The value, \(x\), of \(X\) is determined as follows. If the result of the toss is heads then \(x= \vert ks -1\vert\), and if the result of the toss is tails then \(x=\vert k-s\vert\), where \(s\) is the number on the die and \(k\) is a given number. Show that \(\mathbb{E}(X^2) = k +13(k-1)^2 /6\). Given that both \(\mathbb{E}(X^2)\) and \(\mathbb{E}(X)\) are positive integers, and that \(k\) is a single-digit positive integer, determine the value of \(k\), and write down the probability distribution of \(X\). A gambler pays \(\pounds 1\) to play the game, which consists of two rounds. The gambler is paid:
Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)
For positive integers \(n\), \(a\) and \(b\), the integer \(c_r\) (\(0\le r\le n\)) is defined to be the coefficient of \(x^r\) in the expansion in powers of \(x\) of \((a+bx)^n\). Write down an expression for \(c_r\) in terms of \(r\), \(n\), \(a\) and \(b\). For given \(n\), \(a\) and \(b\), let \(m\) denote a value of \(r\) for which \(c_r\) is greatest (that is, \(c_m \ge c_r\) for \(0\le r\le n\)). Show that \[ \frac{b(n+1)}{a+b} - 1 \le m \le \frac {b(n+1)}{a+b} \,. \] Deduce that \(m\) is either a unique integer or one of two consecutive integers. Let \(G(n,a,b)\) denote the unique value of \(m\) (if there is one) or the larger of the two possible values of \(m\).
Solution: \(c_r = \binom{n}{r}a^{n-r}b^r\) \begin{align*} && c_m &\geq c_{m+1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\ \Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\ \Rightarrow && (m+1)a &\geq (n-m)b \\ \Rightarrow && m(a+b) &\geq nb -a \\ \Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\ \\ && c_m &\geq c_{m-1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\ \Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\ \Rightarrow && (n-m+1)b &\geq ma \\ \Rightarrow && (n+1)b &\geq m(a+b) \\ \Rightarrow && m &\leq \frac{(n+1)b}{a+b} \end{align*} Since \(m\) lies between two values \(1\) apart is is either equal to one of those two values or is the unique integer between them. Let \(\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor\), so
Let \(k\) be an integer satisfying \(0\le k \le 9\,\). Show that \(0\le 10k-k^2\le 25\) and that \(k(k-1)(k+1)\) is divisible by \(3\,\). For each \(3\)-digit number \(N\), where \(N\ge100\), let \(S\) be the sum of the hundreds digit, the square of the tens digit and the cube of the units digit. Find the numbers \(N\) such that \(S=N\). [Hint: write \(N=100a+10b+c\,\) where \(a\,\), \(b\,\) and \(c\) are the digits of \(N\,\).]
Solution: First note that \(10k - k^2 = 25 - (5-k)^2\) which is clearly bounded above by \(25\). The smallest it can be is when \(|5-k|\) is as large as possible, ie when \(k =0\) and we get a lower bound of \(0\). For \((k-1)k(k+1)\) notice this is the product of \(3\) consecutive integers, and therefore must be divisible by \(3\). (In fact, it's divisible by six, since \(\binom{k+1}{3}\) is the number of ways to choose \(3\) objects from \(k+1\). Let \(N = 100a + 10b + c\) where \(0 \leq a,b,c < 10\) and \(1 \leq a\). \(S = a + b^2 + c^3\) we want to find \begin{align*} && 100a +10b + c &= a + b^2 + c^3 \\ \Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\ \end{align*} Therefore \(3 \mid 10b - b^2 = b(10-b)\). Therefore \(3 \mid b\) or \(3 \mid 10-b\) so \(b = 0, 3, 6, 1, 4, 7\) We also have \(99a \geq 99\) and \(10b-b^2 \in [0, 25]\) so we need \(c^3-c \geq 99\), so \(c \geq 5\) Case \(c = 5\), Then \(c^3-c = 120\) so \(a = 1\) and \(10b-b^2 = 21\) so \(b= 3, 7\) \(N = 135, 175\) Case \(c = 6\), so \(c^3 - c = 210\) so \(a = 2\) and \(25-(5-k)^2 = 12\) so no solutions. Case \(c = 7\), so \(7^3 - 7 = 336\) so \(a = 3\) and \(25-(5-k)^2 = 39\) so no solutions. Case \(c = 8\) so \(8^3-8 = 504\) so \(a = 5\) and \(25-(5-k)^2 = 9\), so \(b = 1, 9\) and \(N = 518, 598\) Case \(c = 9\) so \(9^3 - 9 = 720\), so \(a = 7\) and \(25-(5-k)^2 = 27\) so no solutions. Therefore all the solutions are \(N = 135, 175, 518, 598\)
I have two dice whose faces are all painted different colours. I number the faces of one of them \(1,2,2,3,3,6\) and the other \(1,3,3,4,5,6.\) I can now throw a total of 3 in two different ways using the two number \(2\)'s on the first die once each. Show that there are seven different ways of throwing a total of 6. I now renumber the dice (again only using integers in the range 1 to 6) with the results shown in the following table \noindent
| Total shown by the two dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Different ways of obtaining the total | 0 | 2 | 1 | 1 | 4 | 3 | 8 | 6 | 5 | 6 | 0 |
The random variables \(X\) and \(Y\) take integer values \(x\) and \(y\) respectively which are restricted by \(x\geqslant1,\) \(y\geqslant1\) and \(2x+y\leqslant2a\) where \(a\) is an integer greater than 1. The joint probability is given by \[ \mathrm{P}(X=x,Y=y)=c(2x+y), \] where \(c\) is a positive constant, within this region and zero elsewhere. Obtain, in terms of \(x,c\) and \(a,\) the marginal probability \(\mathrm{P}(X=x)\) and show that \[ c=\frac{6}{a(a-1)(8a+5)}. \] Show that when \(y\) is an even number the marginal probability \(\mathrm{P}(Y=y)\) is \[ \frac{3(2a-y)(2a+2+y)}{2a(a-1)(8a+5)} \] and find the corresponding expression when \(y\) is off. Evaluate \(\mathrm{E}(Y)\) in terms of \(a\).