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The probability distribution for the number of eggs in a clutch is \(\operatorname{Po}(\lambda)\), and the probability that each egg will hatch is \(p\) (independently of the size of the clutch). Show by direct calculation that the probability distribution for the number of chicks that hatch is \(\operatorname{Po}(\lambda p)\).

Short Answer

Expert verified
The probability distribution for the number of hatched chicks is given by the Poisson distribution \(\operatorname{Po}(\lambda p)\).

Step by step solution

01

Understand the Given Information

The exercise provides two pieces of information: the Poisson distribution that represents the number of eggs in a clutch, denoted as \(\operatorname{Po}(\lambda)\), and the independent probability of each egg hatching, represented as \(p\). Our goal is to find the probability distribution for the number of hatching chicks.
02

Calculating the Theoretical Probability Distribution

Let's denote \(X\) as the number of eggs in a clutch, and \(Y\) as the number of hatchlings. As \(X\) is Poisson-distributed and since each egg hatches independently with a probability \(p\), the number of chicks that successfully hatch, \(Y\), can be calculated from \(X\), using the formula for the probability of a Poisson-distributed event: \(P(X=k) = \frac{e^{-\lambda} * \lambda^k}{k!}\). Additionally, since each egg hatches independently, \(Y\) then becomes a binomial distribution within the Poisson distribution, so \(P(Y=j|X=k) = \binom{k}{j} * p^j * (1-p)^{k-j}\).
03

Deriving the overall hatching probability distribution

To calculate the overall probability of observing a defined number of hatched chicks, we sum over all the possible values of \(X\): \(P(Y=j) = \sum_{k=0}^{\infty} P(Y=j|X=k)*P(X=k)= \sum_{k=0}^{\infty} \binom{k}{j} * p^j * (1-p)^{k-j} * \frac{e^{-\lambda} * \lambda^k}{k!}\). Simplifying this sum, taking out invariant terms: \(P(Y=j) = e^{-\lambda} * p^j /(j!) * \sum_{k=j}^{\infty} \frac{(1-p)^{k-j}*\lambda^k}{(k-j)!} = e^{-\lambda} * (\lambda p)^j /(j!) * \sum_{k=j}^{\infty} \frac{((1-p)\lambda)^{k-j}}{(k-j)!}\). Using the definition of a Poisson distribution, the summation becomes equal to 1, yielding the final result: \(P(Y=j) = \frac{e^{-\lambda p} * (\lambda p)^j}{j!}\), showing that \(Y\) follows a Poisson distribution \(\operatorname{Po}(\lambda p)\).

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Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that describes the likelihood of different outcomes in a random experiment. In essence, it provides a list of the probabilities associated with each possible outcome. The key to understanding probability distributions is to recognize that they come in two main varieties: discrete and continuous.

Discrete probability distributions, like the Poisson and Binomial distributions, apply to scenarios where the outcomes can be counted individually and are separated. These distributions are characterized by a finite or countably infinite set of outcomes, such as the number of eggs in a nest or the number of heads when flipping a coin multiple times.

Continuous probability distributions, on the other hand, are used when the outcomes form a continuum and can take on any value within a range, such as measurements of height or weight. A common continuous distribution is the normal distribution, which is symmetric and describes many natural phenomena.

To understand and calculate probabilities for different scenarios, one must first determine which type of probability distribution is appropriate. Then, using the properties and formulas associated with that distribution, specific probabilities can be calculated.
Poisson Process
The Poisson process is a stochastic process that serves as a mathematical model for events occurring randomly over time or space. It assumes that these events happen independently of each other and at a constant average rate. The classic example of a Poisson process is radioactive decay, where the decay events are random and independent.

The Poisson distribution arises from this process when counting the number of events occurring within a fixed interval. It's parameterized by \(\lambda\), which corresponds to the average number of occurrences in the interval. Notably, the Poisson distribution is suitable for modeling scenarios with a large number of trials and a small probability of success in each trial, making it different from the Binomial distribution which is better suited to smaller numbers of trials.

In the context of the exercise, we assume a Poisson process for the number of eggs laid, because the eggs appear one by one over time. The eggs hatching can also be seen as part of a Poisson process, where each hatching event is independent and occurs with a fixed probability.
Binomial Distribution
The Binomial distribution is another key type of discrete probability distribution. It is used to model the number of successes in a fixed number of independent trials, with the same probability of success on each trial. This distribution is defined by two parameters: the number of trials (n) and the probability of success (p).

Lean on the classic example of flipping a coin, which we assume is fair, the probability of getting heads (success) is \(p=0.5\). If we flip the coin 10 times, the Binomial distribution can predict the probability of getting exactly 6 heads. The specific formula for the probability of k successes out of n trials is given by \(P(X=k) = \binom{n}{k} * p^k * (1-p)^{n-k}\), where \(\binom{n}{k}\) is a binomial coefficient.

In the given exercise, each egg represents an independent trial with probability p of hatching (success). Thus, if we just consider the number of eggs in a single clutch, we could model the number of hatched chicks using a Binomial distribution. However, since the number of eggs itself is random and follows a Poisson distribution, the distribution of hatched chicks is instead shown to be Poisson with parameter \(\lambda p\).

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Most popular questions from this chapter

This problem shows that the odds are hardly ever 'evens' when it comes to dice rolling. (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?

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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 \(X\) ?

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