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Chapter 26: Problem 16

Kittens from different litters do not get on with each other and fighting breaks out whenever two kittens from different litters are present together. A cage initially contains \(x\) kittens from one litter and \(y\) from another. To quell the fighting, kittens are removed at random, one at a time, until peace is restored. Show, by induction, that the expected number of kittens finally remaining is $$ N(x, y)=\frac{x}{y+1}+\frac{y}{x+1} $$

Short Answer

Expert verified
The expected number of kittens finally remaining is \(\frac{x}{y+1} + \frac{y}{x+1}\)

Step by step solution

01

- Base Case

Verify the base case for the induction. When one of the litters is empty, either when \(y = 0\) or \(x = 0\): For \(N(x, 0)\), the expected number of kittens remaining is \(x\). When \(y = 0\), the formula becomes \(N(x, 0) = \frac{x}{0 + 1} + \frac{0}{x + 1} = x\), which matches the expectation as there are no kittens from another litter to cause fighting.
02

- Inductive Hypothesis

Assume the formula holds for \((x, y)\): \[N(x, y) = \frac{x}{y+1} + \frac{y}{x+1}\]Now, consider the case of \((x+1, y)\) and \((x, y+1)\).
03

- Add a kitten to the (x) Litter

Consider the case where an additional kitten is added to the litter with \(x\) kittens, making it \(x+1\): According to our hypothesis:Expected number of kittens remaining in this case can be represented as: \[N(x+1, y) = \frac{x+1}{y+1} + \frac{y}{x+2}\]Which simplifies to adding one more kitten to the litter and adjusting the expectation proportionately.
04

- Add a kitten to the (y) Litter

Consider the case where an additional kitten is added to the litter with \(y\) kittens making it \(y+1\):According to our hypothesis:Expected number of kittens remaining in this case can be represented as: \[N(x, y+1) = \frac{x}{y+2} + \frac{y+1}{x+1}\]This accounts for increased competition due to added kittens, thus the expected outcome conforms to our hypothesis for all general cases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value, often referred to as the mean, is a fundamental concept in probability and statistics. It represents the average outcome one would anticipate from an experiment if it were repeated many times. Here, the expected value helps us understand the average number of kittens remaining after repeated random removals.

In the given problem, we use the formula: \(N(x, y) = \frac{x}{y+1} + \frac{y}{x+1}\) to determine the expected number of kittens remaining when peace is restored. This formula takes into account the competition between the two litters.

**Key Points:**
  • The expected value provides a sense of the average outcome.
  • In our context, it informs us about the average number of kittens left after removing them one by one to stop the fighting.
  • The formula helps predict outcomes in a probabilistic scenario.
Understanding the expected value is crucial as it assists in making predictions about future events based on past occurrences.
Probability
Probability is the measure of the likelihood that a particular event will occur. It ranges from 0 (the event will not happen) to 1 (the event will definitely happen). In the given problem, probability helps us determine how likely it is for a specific number of kittens to remain after repeatedly removing them one by one.

**Key Components:**
  • Events: These are outcomes or occurrences that can be counted.
  • Probability Distribution: This shows all possible outcomes and their associated probabilities.
In our scenario, the probability distribution could describe the different possible numbers of kittens remaining and the likelihood of each of those numbers.
For instance, if you know the probability of each kitten being removed, you can calculate the probability of different outcomes for the number of kittens left in the cage. This understanding allows for a probabilistic model of the kitten problem, ensuring precise and reliable predictions.
Combinatorial Problems
Combinatorial problems involve counting and arranging different possible outcomes. They often require finding the number of ways to choose or arrange items, considering various constraints. In the given problem, combinatorics helps in understanding different possible sequences of kitten removals and their outcomes.

**Key Concepts in Combinatorics:**
  • Permutations: These are different arrangements of a set of items.
  • Combinations: These are selections of items without regard to order.
  • Counting Principles: Principles like the addition and multiplication rules help in solving combinatorial problems.
Understanding combinatorial methods allows us to count the number of ways kittens can be removed without reintroducing fighting, thus leading to the final expected number. It helps break down the problem into manageable parts and identify patterns or sequences that align with the expectations established by the formula.
Utilizing these methods, we can confirm the consistency and correctness of our predictions within the proposed hypotheses and refined calculations.

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Most popular questions from this chapter

A set of \(2 N+1\) rods consists of one of each integer length \(1,2, \ldots, 2 N, 2 N+1\) Three, of lengths \(a, b\) and \(c\), are selected, of which \(a\) is the longest. By considering the possible values of \(b\) and \(c\), determine the number of ways in which a nondegenerate triangle (i.e. one of non-zero area) can be formed (i) if \(a\) is even, and (ii) if \(a\) is odd. Combine these results appropriately to determine the total number of non-degenerate triangles that can be formed with the \(2 N+1\) rods, and hence show that the probability that such a triangle can be formed from a random selection (without replacement) of three rods is $$ \frac{(N-1)(4 N+1)}{2\left(4 N^{2}-1\right)} $$

A shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the final digit (the number of pence) in the price is uniformly and randomly distributed from 0 to \(9 .\) Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probability if the price ends in a ' 5 '. Should she suspect a mistake if the cashier asks her for 23 pence more than she estimated?

(a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in \(A\) 's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(£ 1.10\) per game, as compared to \(D\) 's \(£ 1.00\) stake. Who will show a profit?

\(26.3 A\) and \(B\) each have two unbiased four-faced dice, the four faces being numbered \(1,2,3,4\). Without looking, \(B\) tries to guess the sum \(x\) of the numbers on the bottom faces of \(A\) 's two dice after they have been thrown onto a table. If the guess is correct \(B\) receives \(x^{2}\) euros, but if not he loses \(x\) euros. Determine \(B\) 's expected gain per throw of \(A\) 's dice when he adopts each of the following strategies: (a) he selects \(x\) at random in the range \(2 \leq x \leq 8\); (b) he throws his own two dice and guesses \(x\) to be whatever they indicate; (c) he takes your advice and always chooses the same value for \(x\). Which number would you advise? 26.4 Use the method of induction to prove equation (26.16), the probability addition law for the union of \(n\) general events.

For a non-negative integer random variable \(X\), in addition to the probability generating function \(\Phi_{X}(t)\) defined in equation (26.71) it is possible to define the probability generating function $$ \Psi_{X}(t)=\sum_{n=0}^{\infty} g_{n} t^{n} $$ where \(g_{n}\) is the probability that \(X>n\). (a) Prove that \(\Phi_{X}\) and \(\Psi_{X}\) are related by $$ \Psi_{X}(t)=\frac{1-\Phi_{X}(t)}{1-t} $$ (b) Show that \(E[X]\) is given by \(\Psi_{X}(1)\) and that the variance of \(X\) can be expressed as \(2 \Psi_{X}^{\prime}(1)+\Psi_{X}(1)-\left[\Psi_{X}(1)\right]^{2}\) (c) For a particular random variable \(X\), the probability that \(X>n\) is equal to \(\alpha^{n+1}\) with \(0<\alpha<1\). Use the results in \((\mathrm{b})\) to show that \(V[X]=\alpha(1-\alpha)^{-2}\).
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