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Chapter 3: Problem 77

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes \(1,2,\) or \(3 .\) Given that outcome 3 is the last of the three outcomes to occur, find the conditional probability that. (a) the first trial results in outcome 1. (b) the first two trials both result in outcome \(1 .\)

Short Answer

Expert verified
Given that outcome 3 is the last among outcomes 1, 2, and 3 to occur, the probabilities are as follows: (a) The probability that the first trial results in outcome 1: P(B1|A) = 1/2. (b) The probability that the first two trials both result in outcome 1: P(B2|A) = 1/2.

Step by step solution

01

1. Find P(A)

The probability of outcome 3 being the last among outcomes 1, 2, and 3 is always 1/3 since these three outcomes are equally likely.
02

2. Find P(B1)|A

Given that outcome 3 is the last among outcomes 1, 2, and 3 to occur, we want to find the probability that the first trial results in outcome 1. Before outcome 3 occurs, one of the other two outcomes (1 or 2) must occur first. Since they are equally likely, the probability of having outcome 1 in the first trial is 1/2. We don't need to consider the case when outcome 3 occurs at the first trial, because we're given the condition A: P(B1|A) = P(B1|Outcome 3 is the last in the sequence) = 1/2
03

3. Find P(B2|A)

Given that outcome 3 is the last among outcomes 1, 2, and 3 to occur, we want to find the probability that the first two trials both result in outcome 1. Let's define a new event: C: The first trial results in outcome 1 and the second trial results in outcome 1. Now, we need to find the probability of event C happening given event A. Since we already know that the first trial has to result in outcome 1 (i.e., P(B1|A) = 1/2), we don't need to consider the case when outcome 3 occurs in the first trial: P(B2|A) = P(C|A) = P(First two trials result in outcome 1 | Outcome 3 is the last in the sequence) Before outcome 3 occurs, one of the other two outcomes (1 or 2) must occur first. To find the probability of having the first two trials resulting in outcome 1, we need to find the probability of having outcome 1 in the second trial, given outcome 1 in the first trial and outcome 3 is the last in the sequence. Since the trials' outcomes are independent and before outcome 3 occurs, the outcomes 1 and 2 are equally likely, we have: P(B2|A) = P(C|Outcome 1 in the first trial and Outcome 3 is the last in the sequence) = 1/2 #Final Answer# (a) P(B1|A) = 1/2 (b) P(B2|A) = 1/2

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Key Concepts

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

Independent Trials
In probability theory, independent trials are a set of experiments or events where the outcome of one trial doesn't affect the outcome of another. Each trial is isolated in terms of its result. This is crucial when considering conditional probabilities, as it allows us to analyze each event separately without worry about interactions between them.

For our exercise, each trial in the sequence is independent. This means the result of any one trial (such as getting outcome 1, 2, or 3) does not change how the results will be in the following trials. Because of this independence, when calculating the probability of multiple trials, we multiply the probabilities of outcomes in each individual trial. This is possible only when trials are independent.

Understanding independence helps students realize that outcomes remain fair and consistent across trials, simplifying the calculations as seen in determining the outcomes of the first or first two trials.
Equally Likely Outcomes
When outcomes are equally likely, each outcome has the same probability of occurring in a given event. This concept is essential in understanding fair situations in probability.

For instance, in our exercise, outcomes 1, 2, and 3 are equally likely to happen in each trial. This means that for any single trial, the probability of obtaining 1, 2, or 3 is equal, which is \(\frac{1}{3}\). Having outcomes that are equally probable simplifies calculations significantly.

Knowing all outcomes are equally likely ensures fairness and balance in the scenario. It lets us confidently state that without any external factors, all potential results hold an equal chance, an assumption that is widely used when setting up probability problems.
Probability Calculation
Calculating probability helps us quantify how likely an event is to occur. For equally likely outcomes in independent trials, the probability of an event happening can be straightforward. However, when we introduce conditions, things get slightly more complex.

In the exercise provided, we need to find the conditional probabilities. Conditional probability refers to the probability of an event occurring given that another event has already occurred.
  • To find the probability that the first trial results in outcome 1, given that outcome 3 is last, we focus only on trials before outcome 3. This gives us: \(P(B1|A) = \frac{1}{2}\).
  • Similarly, for the first two trials to both be outcome 1, while ensuring outcome 3 is the last, we also find: \(P(B2|A) = \frac{1}{2}\).
The key here is understanding how the condition "outcome 3 is the last to occur" influences what earlier trials can do, emphasizing the role conditions play in these calculations.
Sequence of Events
The sequence of events in probability involves the arrangement or order in which these events occur. How this sequence plays out can substantially impact calculations, especially with conditions or restrictions.

For our exercise, the sequence of obtaining outcome 3 last imposes conditions on how earlier trials can unfold. This order specifically affects our approach to calculating probabilities for trials before outcome 3 appears.

Each trial in the sequence is crucial. By understanding that outcome 3 won't appear before others, we can focus purely on the interactions between outcomes 1 and 2. In essence, this restriction simplifies our problem by reducing uncertainty around the later appearance of outcome 3, guiding accurate probability assessments for earlier outcomes.

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Most popular questions from this chapter

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair \(x X\) then it will outwardly have the appearance of the \(X\) gene. For instance, if \(X\) stands for brown eyes and \(x\) for blue eyes, then an individual having either gene pair \(X X\) or \(x X\) will have brown eyes, whereas one having gene pair \(x x\) will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2 organisms with respective genotypes \(a A, b B, c c, d D\) ee and \(A A, B B, c c, D D,\) ee would have different genotypes but the same phenotype.) In a mating between 2 organisms, each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism's mate. In a mating between organisms having genotypes \(a A, b B, c C, d D, e E\) and \(a a, b B, c c\) \(D d,\) ee what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent? (b) the second parent? (c) either parent? (d) neither parent?

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