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Chapter 3: Problem 90

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 outcomes 1 and 2 both occur at least once in \(n\) independent trials is: \[P(\text{1 and 2 occur at least once}) = 1 - [(p_0 + p_2)^n + (p_0 + p_1)^n - p_0^n]\]

Step by step solution

01

Find the probability of the complementary events

To begin, let's first find the probability that outcome 1 never occurs. In this case, the only possible outcomes for each trial are 0 and 2. The probability of this happening in a single trial is \(p_0 + p_2\). Since we have \(n\) independent trials, we raise this probability to the power of n: \((p_0 + p_2)^n\). Similarly, the probability that outcome 2 never occurs is \((p_0 + p_1)^n\). Finally, the probability that both outcome 1 and outcome 2 never occur is \(p_0^n\).
02

Apply complementary event principle and sum probabilities

Now, let's use the complementary event principle. The probability that outcome 1 or outcome 2 never occur is the sum of the individual probabilities, minus the probability that both outcome 1 and outcome 2 never occur (that is, the intersection of the events): \[P(\text{complementary}) = (p_0 + p_2)^n + (p_0 + p_1)^n - p_0^n\]
03

Calculate the desired probability

Finally, to find the probability that both outcomes 1 and 2 occur at least once, we will subtract the probability of the complementary event from 1: \[P(\text{1 and 2 occur at least once}) = 1 - P(\text{complementary}) = 1 - [(p_0 + p_2)^n + (p_0 + p_1)^n - p_0^n]\] The probability that outcomes 1 and 2 both occur at least once in \(n\) independent trials is: \[P(\text{1 and 2 occur at least once}) = 1 - [(p_0 + p_2)^n + (p_0 + p_1)^n - p_0^n]\]

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Key Concepts

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

Complementary Events
In probability theory, complementary events refer to pairs of dichotomous outcomes where the occurrence of one event implies the non-occurrence of the other and vice versa. For example, if you flip a coin, the result could either be heads or tails; these are complementary events. If one happens, the other can't.

When dealing with complementary events, it might be easier to calculate the probability of an event by using its complement. Because the sum of probabilities for all possible outcomes equals 1, the probability of an event happening is 1 minus the probability of it not happening.

In the given problem, we calculated the probability of the complementary event, which is when outcomes 1 and 2 do not occur at least once. This involves calculating each of the individual probabilities of not having 1 or not having 2, and subtracting the overlap (probability of neither 1 nor 2).

This approach simplifies calculations and avoids the direct computation of increasingly complex scenarios, especially when independent trials are involved.
Independent Trials
Independent trials represent situations where the outcome of one trial doesn't affect the outcome of another. Each trial is like hitting reset; no matter what happened before, the odds remain the same for each new attempt.

In this problem, the assumption is that each trial, which can result in outcome 0, 1, or 2, is independent. This independence allows us to multiply probabilities when calculating the likelihood of events over multiple trials.
  • To find the probability of a specific sequence of outcomes, multiply their individual probabilities.
  • In series of trials, if the probability for one outcome stays constant, it's an indication of independence.
This concept plays a crucial role in computing probabilities across many scenarios, ensuring that cumulative calculations account for repeated, unchanged conditions.
Conditional Probability
Conditional probability is used to find the likelihood of an event occurring, given that another event has already happened. It's how probability adapts when we're informed by additional data.

Mathematically, conditional probability, where we want to find the probability of event A given event B has occurred, is usually written as:\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
In our exercise, conditional probability isn't directly applied, but understanding it helps assimilate complementary events. It provides insight into how probabilities adjust when certain conditions are established. Though our problem is solved by complementarity and independence, knowing when events aren't independent would lead us to use this concept, changing how we calculate related probabilities. Understanding these foundational tools can expand your reasoning in probability and enhance problem-solving skills efficiently.

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Most popular questions from this chapter

All the workers at a certain company drive to work and park in the company's lot. The company is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Explain your answer. 1\. Randomly choose \(n\) workers, find out how many were in the cars in which they were driven, and take the average of the \(n\) values. 2\. Randomly choose \(n\) cars in the lot, find out how many were driven in those cars, and take the average of the \(n\) values.

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin which lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2} .\) Consider the following procedure for accomplishing our task: 1\. Flip the coin. 2\. Flip the coin again. 3\. If both flips land on heads or both land on tails, return to step 1. 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

An engineering system consisting of \(n\) components is said to be a \(k\) -out- of- \(n\) system \((k \leq n)\) if the system functions if and only if at least \(k\) of the \(n\) components function. Suppose that all components function independently of one another. (a) If the \(i\) th component functions with probability \(P_{i}, i=\) \(1,2,3,4,\) compute the probability that a 2 -out-of- 4 system functions. (b) Repeat part (a) for a 3 -out-of- 5 system. (c) Repeat for a \(k\) -out-of- \(n\) system when all the \(P_{i}\) equal \(p\) (that is, \(\left.P_{i}=p, i=1,2, \ldots, n\right)\)

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 dic \(A ;\) if it lands on tails, then dic \(B\) 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?

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly. (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.
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