Problem 49
Question
Suppose the probability that an event will occur in one trial is \(p .\) Show that the probability that the event will occur at least once in \(n\) independent trials is \(1-(1-p)^{n} .\)
Step-by-Step Solution
Verified Answer
The probability of the event occurring at least once in \(n\) independent trials is given by: \[1 - (1-p)^n.\]
1Step 1: Complementary probability principle
Since we want to find the probability of the event occurring at least once in \(n\) independent trials, it's easier to first determine the probability of the event NOT occurring at all in \(n\) trials, and then use the complementary probability principle.
2Step 2: Find the probability of the event not occurring in a single trial
Given that the probability of the event occurring in a single trial is \(p\), the probability that it does NOT occur in a single trial is \(1-p\).
3Step 3: Find the probability of the event not occurring in all \(n\) trials
Since the trials are independent, the probability of the event not occurring in all \(n\) trials is the product of the probabilities of it not occurring in each trial. So, the probability of the event not occurring in all \(n\) trials is \((1-p)^n\).
4Step 4: Apply the complementary probability principle
Finally, apply the complementary probability principle to find the probability of the event occurring at least once in \(n\) trials. The probability that the event occurs at least once in \(n\) trials is 1 minus the probability of it not occurring in all \(n\) trials.
Thus, the probability of the event occurring at least once in \(n\) trials is given by: \[1 - (1-p)^n.\]
Key Concepts
Independent EventsComplementary ProbabilityBinomial Probability
Independent Events
In probability theory, independent events are two or more events where the occurrence of one does not affect the other. If you have multiple trials or experiments, such as flipping a coin, rolling a die, or drawing cards, each act is independent if the outcome of one does not influence the next.
To understand independence in the context of our exercise, assume you are repeating a trial multiple times. Let's say you are rolling a fair die. Rolling a 4 on the first try does not change the probability of rolling a 4 on the second try—each roll is independent of the others.
Mathematically, for two events A and B, they are independent if \[ P(A \cap B) = P(A) \times P(B) \]. For our exercise, since the trials are independent, the probability of the event not occurring in all \( n \) trials is calculated by multiplying the probability of it not happening in each trial, leading to \( (1-p)^n \).
To understand independence in the context of our exercise, assume you are repeating a trial multiple times. Let's say you are rolling a fair die. Rolling a 4 on the first try does not change the probability of rolling a 4 on the second try—each roll is independent of the others.
Mathematically, for two events A and B, they are independent if \[ P(A \cap B) = P(A) \times P(B) \]. For our exercise, since the trials are independent, the probability of the event not occurring in all \( n \) trials is calculated by multiplying the probability of it not happening in each trial, leading to \( (1-p)^n \).
Complementary Probability
The concept of complementary probability is an efficient tool for calculating the likelihood of an event happening at least once. In simple terms, for any event A, there is a complement, denoted by \( A' \), which represents the event A not occurring.
The probability of the complement of an event is given by \[ P(A') = 1 - P(A) \]. In the context of our exercise, we wanted to compute the probability of an event happening at least once over \( n \) trials.
To do this, it was easier first to calculate the probability of it not occurring in all trials, \( (1-p)^n \), and then subtracting that from 1. This gives \[ P \text{(at least one occurrence in n trials)} = 1 - (1-p)^n \]. This way of thinking simplifies problems and reduces calculation errors.
The probability of the complement of an event is given by \[ P(A') = 1 - P(A) \]. In the context of our exercise, we wanted to compute the probability of an event happening at least once over \( n \) trials.
To do this, it was easier first to calculate the probability of it not occurring in all trials, \( (1-p)^n \), and then subtracting that from 1. This gives \[ P \text{(at least one occurrence in n trials)} = 1 - (1-p)^n \]. This way of thinking simplifies problems and reduces calculation errors.
Binomial Probability
Binomial probability relates to the probability of a specific number of successes in a series of independent and identically-repeated trials, each with the same probability of success.
The classic example is flipping a fair coin, where the probability of heads (success) is 0.5, repeated across a number of flips (trials). For such scenarios, binomial distribution provides a powerful model that captures the essence of a sequence of successes over trials.
The formula for binomial probability is \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is a binomial coefficient, \( p \) is the probability of success, and \( n \) is the number of trials.
Although the exercise at hand doesn't directly use the binomial probability formula, understanding that the event occurring at least once is part of a binomial distribution clarifies that even rare events ( small \( p \)) have a non-zero chance of happening over multiple trials.
The classic example is flipping a fair coin, where the probability of heads (success) is 0.5, repeated across a number of flips (trials). For such scenarios, binomial distribution provides a powerful model that captures the essence of a sequence of successes over trials.
The formula for binomial probability is \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is a binomial coefficient, \( p \) is the probability of success, and \( n \) is the number of trials.
Although the exercise at hand doesn't directly use the binomial probability formula, understanding that the event occurring at least once is part of a binomial distribution clarifies that even rare events ( small \( p \)) have a non-zero chance of happening over multiple trials.
Other exercises in this chapter
Problem 48
According to the Centers for Disease Control and Prevention, the percentage of adults 25 yr and older who smoke, by educational level, is as follows: $$\begin{a
View solution Problem 48
Let \(E\) be any event in a sample space \(S .\) a. Are \(E\) and \(S\) independent? Explain your answer. b. Are \(E\) and \(\varnothing\) independent? Explain
View solution Problem 51
Let \(E\) and \(F\) be events such that \(F \subset E\). Find \(P(E \mid F)\) and interpret your result.
View solution Problem 52
Suppose that \(A\) and \(B\) are mutually exclusive events and that \(P(A \cup B) \neq 0 .\) What is \(P(A \mid A \cup B)\) ?
View solution