/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 Suppose that \(n\) independent t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(n\) independent trials, each of which results in any of the outcomes \(0,1,\) or \(2,\) with respective probabilities \(p_{0}, p_{1},\) and \(p_{2}, \sum_{i=0}^{2} p_{i}=1,\) are performed. Find the probability that outcomes 1 and 2 both occur at least once.

Short Answer

Expert verified
The probability that both outcomes 1 and 2 occur at least once in the given n independent trials is: \[1 - [(p_0 + p_2)^n + (p_0 + p_1)^n]\]

Step by step solution

01

Probability of not having outcome 1 or outcome 2 in the trials

First, we need to calculate the probability that either outcome 1 or outcome 2 doesn't occur at all in the n trials. For this, we have two scenarios to consider: 1. Outcome 1 doesn't occur, which means that all the trials have either outcome 0 or outcome 2, with probabilities \(p_0\) and \(p_2\) respectively. 2. Outcome 2 doesn't occur, which means that all the trials have either outcome 0 or outcome 1, with probabilities \(p_0\) and \(p_1\) respectively.
02

Calculate the probability of each scenario

Now we will calculate the probability of each one of the two scenarios: 1. The probability of outcome 1 not occurring at all: Since we only have outcomes 0 and 2 in the n trials, the combined probability of these two outcomes is \(p_0 + p_2\). Since the trials are independent, the probability of this scenario happening is \((p_0 + p_2)^n\). 2. The probability of outcome 2 not occurring at all: Similarly, since we only have outcomes 0 and 1 in the n trials, the combined probability of these two outcomes is \(p_0 + p_1\). The probability of this scenario happening is \((p_0 + p_1)^n\).
03

Calculate the probability of both scenarios

Since both of the scenarios are mutually exclusive, we can calculate the probability of any of them happening by simply adding their individual probabilities: The probability of either outcome 1 or outcome 2 not occurring at all is \((p_0 + p_2)^n + (p_0 + p_1)^n\).
04

Calculate the probability of the complementary event

Now we can find the probability of the complementary event i.e., both outcomes 1 and 2 occur at least once. As both events are complementary, the probability of one is equal to 1 minus the probability of the other: The probability of both outcomes 1 and 2 occurring at least once is \(1 - [(p_0 + p_2)^n + (p_0 + p_1)^n]\). So, the final result for the probability that both outcomes 1 and 2 occur at least once is: \[1 - [(p_0 + p_2)^n + (p_0 + p_1)^n]\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Independent Events
In probability theory, independent events are situations where the occurrence of one event does not affect the probability of another. Imagine flipping a coin or rolling a dice; the result of one has no impact on the subsequent flips or rolls. These are independent instances.
  • Mathematically, events are independent if the probability of both occurring, denoted as \( P(A \cap B) \), equals the product of their individual probabilities: \( P(A) \times P(B) \).
  • This concept is crucial for understanding complex probabilities since it simplifies calculations.

In the exercise, each trial's result is independent, meaning the outcome of one trial (0, 1, or 2) doesn't influence the results of any other trial. This independence allows us to calculate combined probabilities by multiplying the likelihoods of each event. For instance, if each trial independently has probabilities \(p_0, p_1,\) and \(p_2\) for outcomes 0, 1, and 2 respectively, this independence helps us deduce the overall probability of specific event combinations over multiple trials.
Complementary Events
Complementary events are pairs of events where one event occurs if and only if the other does not. The concept of a complementary event is straightforward if you think about common dichotomies like heads and tails in a coin flip.
  • If event A is the occurrence of heads, then its complement, event A', is the occurrence of tails.
  • For any event A, the sum of the probabilities of A and its complement is 1: \( P(A) + P(A') = 1 \).

This principle is used in the given exercise when calculating the probability of both outcomes 1 and 2 occurring at least once. The formula \[ 1 - [(p_0 + p_2)^n + (p_0 + p_1)^n] \] utilizes the concept of complementary events. Here, the probability of neither outcome 1 nor 2 occurring (at all) is subtracted from 1, giving us the probability that both outcomes occur at least once.
Outcomes in Probability
Outcomes in probability refer to the possible results from an experiment or event. Each possible result is considered an outcome, and the total of all potential outcomes comprises the sample space.
  • Every outcome has an associated probability, a measure ranging from 0 to 1 that indicates how likely that outcome is.
  • In our exercise, the potential outcomes for each trial are 0, 1, or 2, with specific associated probabilities \(p_0, p_1,\) and \(p_2\), respectively.

The sample space in this context consists of these three discrete outcomes. Understanding the distribution of these probabilities is vital, as it affects the computation of more complex probabilities, like the likelihood that outcome 1 and 2 occur at least once over multiple trials. By examining how individual outcomes combine, we can better predict and articulate the specifics of probability scenarios, which can be much more intricate than initially perceived.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade given that the second and third cards are spades.

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if \(A\) knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\) because he would then be one of two prisoners. What do you think of the jailer's reasoning?

If you had to construct a mathematical model for events \(E\) and \(F,\) as described in parts (a) through (e), would you assume that they were independent events? Explain your reasoning. (a) \(E\) is the event that a businesswoman has blue eyes, and \(F\) is the event that her secretary has blue eyes. (b) \(E\) is the event that a professor owns a car, and \(F\) is the event that he is listed in the telephone book. (c) \(E\) is the event that a man is under 6 feet tall, and \(F\) is the event that he weighs more than 200 pounds. (d) \(E\) is the event that a woman lives in the United States, and \(F\) is the event that she lives in the Western Hemisphere. (e) \(E\) is the event that it will rain tomorrow, and \(F\) is the event that it will rain the day after tomorrow.

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die \(A\); if it lands on tails, then die \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}\) (b) If the first two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the first two throws, what is the probability that it is die \(A\) that is being used?

In any given year, a male automobile policyholder will make a claim with probability \(p_{m}\) and a female policyholder will make a claim with probability \(p_{f},\) where \(p_{f} \neq p_{m} .\) The fraction of the policyholders that are male is \(\alpha, 0<\alpha<1 .\) A policyholder is randomly chosen. If \(A_{i}\) denotes the event that this policyholder will make a claim in year \(i,\) show that $$ P\left(A_{2} | A_{1}\right)>P\left(A_{1}\right) $$ Give an intuitive explanation of why the preceding inequality is true.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.