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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

In the game of Blackball, at each turn Muggins draws a ball at random from a bag containing five white balls, three red balls and two black balls; after being recorded, the ball is replaced in the bag. A white ball earns him \(\$ 1\) whilst a red ball gets him \(\$ 2\); in either case he also has the option of leaving with his current winnings or of taking a further turn on the same basis. If he draws a black ball the game ends and he loses all he may have gained previously. Find an expression for Muggins' expected return if he adopts the strategy to drawing up to \(n\) balls if he has not been eliminated by then. Show that, as the entry fee to play is \$3, Muggins should be dissuaded from playing Blackball, but if that cannot be done what value of \(n\) would you advise him to adopt?

An electronics assembly firm buys its microchips from three different suppliers; half of them are bought from firm \(X\), whilst firms \(Y\) and \(Z\) supply \(30 \%\) and \(20 \%\) respectively. The suppliers use different quality- control procedures and the percentages of defective chips are \(2 \%, 4 \%\) and \(4 \%\) for \(X, Y\) and \(Z\) respectively. The probabilities that a defective chip will fail two or more assembly-line tests are \(40 \%, 60 \%\) and \(80 \%\) respectively, whilst all defective chips have a \(10 \%\) chance of escaping detection. An assembler finds a chip that fails only one test. What is the probability that it came from supplier \(X\) ?

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

A continuous random variable \(X\) has a probability density function \(f(x)\); the corresponding cumulative probability function is \(F(x) .\) Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .

A continuous random variable \(X\) is uniformly distributed over the interval \([-c, c]\). A sample of \(2 n+1\) values of \(X\) is selected at random and the random variable \(Z\) is defined as the median of that sample. Show that \(Z\) is distributed over \([-c, c]\) with probability density function, $$ f_{n}(z)=\frac{(2 n+1) !}{(n !)^{2}(2 c)^{2 n+1}}\left(c^{2}-z^{2}\right)^{n} $$ Find the variance of \(Z\).
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