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An event independently occurs on each day with probability \(p .\) Let \(N(n)\) denote the total number of events that occur on the first \(n\) days, and let \(T_{r}\) denote the day on which the \(r\) th event occurs. (a) What is the distribution of \(N(n) ?\) (b) What is the distribution of \(T_{1}\) ? (c) What is the distribution of \(T_{r} ?\) (d) Given that \(N(n)=r\), show that the set of \(r\) days on which events occurred has the same distribution as a random selection (without replacement) of \(r\) of the values \(1,2, \ldots, n\).

Short Answer

Expert verified
The distributions for the given cases are as follows: (a) For the total number of events happening in the first n days: \(N(n) \sim \text{Binomial}(n, p)\). (b) For the day on which the first event occurs: \(T_1 \sim \text{Geometric}(p)\). (c) For the day on which the r-th event occurs: \(T_r=\left\{\begin{array}{ll}0 & \text { if } k<r \\ \binom{k-1}{r-1} p^{r} q^{k-r} & \text { if } k \geq r\end{array}\right.\). (d) Given that N(n) = r, the set of r days on which events occurred has the same distribution as a random selection (without replacement) of r values from 1, 2, ..., n.

Step by step solution

01

(a) Finding the distribution of N(n)#

Since the events occur independently with probability p on each day, we can model the number of events occurring in the first n days as a Binomial distribution. Therefore, the distribution of N(n) is: $$ N(n) \sim \text{Binomial}(n, p) $$ where n is the number of days, and p is the probability of an event occurring on each day.
02

(b) Finding the distribution of T鈧#

The day T鈧 on which the first event occurs is a Geometric distribution because it represents the number of failures before the first success (event) occurs. Therefore, the distribution of T鈧 is: $$ T_1 \sim \text{Geometric}(p) $$ where p is the probability of an event occurring on each day.
03

(c) Finding the distribution of T鈧#

Let q = 1 - p denote the probability of no event happening on a day. For the r-th event to happen on the k-th day, we need (k-r) failed/off days and (r-1) successful event days. The last day k is the r-th event. Therefore, the distribution of T鈧 is given by: $$ T_r=\left\{\begin{array}{ll}0 & \text { if } k<r \\ \binom{k-1}{r-1} p^{r} q^{k-r} & \text { if } k \geq r\end{array}\right. $$
04

(d) Showing that N(n) and random selection have the same distribution#

Given that N(n)=r, the probability of r events occurring in n days can be represented as: $$ P\text{(r events)}=\binom{n}{r} p^{r} q^{n-r} $$ This is the same as selecting r days out of n without replacement, i.e., choosing r elements from a set of n elements. $$ P\text{(choosing r days)}=\frac{\binom{n}{r}}{\binom{n}{r}} = 1 $$ The probability of choosing r days from the 1, 2, ..., n days is the same as the probability of r events occurring in n days; hence their distributions are the same.

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Key Concepts

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

Binomial Distribution
When we talk about the binomial distribution, we are referring to scenarios where there is a fixed number of trials, each with two possible outcomes: success or failure. In our exercise, we have events that occur independently across multiple days. The number of successful occurrences of these events over the first \( n \) days can be modeled by a binomial distribution because each day represents a trial with a success probability of \( p \).
Consequently, the expression \( N(n) \sim \text{Binomial}(n, p) \) neatly encapsulates this, where:
  • \( n \) is the total number of days (or trials),
  • \( p \) is the probability that an event will occur on any given day.
The beauty of the binomial distribution lies in its simplicity and its wide applicability in understanding phenomena ranging from flipping coins to predicting weather patterns.
Geometric Distribution
The geometric distribution is essential for understanding the timing of the first occurrence of an event. In simpler terms, it helps us figure out the likelihood of when something will happen for the first time. In the exercise, \( T_1 \) represents the first day an event occurs.
This scenario is a classic case for a geometric distribution because it models the number of trials until the first success. Here, each day before the event occurs is a 'failure', and the day the event happens is a 'success'. The probability \( T_1 \sim \text{Geometric}(p) \) indicates:
  • Each day has a probability \( p \) that the event will occur.
  • The nature of this distribution means the number of "failures" before the first "success" is being counted.
Geometric distributions are instrumental in scenarios like waiting times, or predicting how often you'll attempt something before succeeding.
Random Selection
Random selection is a concept closely related to probability and involves picking elements from a set such that each element has an equal chance of being chosen. In our exercise, we examine the selection of days on which events occur when \( N(n) = r \).
Given that exactly \( r \) events happen over \( n \) days, these \( r \) days appear as if they were randomly selected from the total days.
This is similar to drawing r elements from a set of n without replacement鈥攅very combination of chosen days is equally probable. This idea reinforces why we can compare it directly to combinatorial selection, emphasizing the equal likelihood of all possible selections, and thus their identical distributions.
Independent Events
Independent events are a foundational concept in probability, where the occurrence of one event does not affect the likelihood of another. In the exercise, the daily occurrence of events is stated to happen independently. This means the outcome on one day doesn't influence the next day's result in any way.
The independence characterizes both the binomial and geometric distributions derived in this scenario:
  • In a binomial setup, each day (or trial) is independent, regardless of what happens on other days.
  • For the geometric distribution, whether an event occurs on one day doesn't affect if it will occur on subsequent days.
This independence is crucial as it allows us to use these probability distributions accurately to model real-world scenarios. Without it, the probability calculations would become much more complex due to interactions between dependent events.

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