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Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1. (b) Explain why \(P(r \geq 1)=1-P(0)\) (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\) (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

Short Answer

Expert verified
Binomial random variable values range from 0 to n, summing to 1. \( P(r \geq 1) = 1 - P(0) \), \( P(r \geq 2) = 1 - P(0) - P(1) \), and in general, \( P(r \geq m) = 1 - P(0) - \cdots - P(m-1) \).

Step by step solution

01

Understand the Sample Space of a Binomial Random Variable

A binomial random variable, say \( r \), represents the number of successes in \( n \) independent trials of a binary experiment (such as flipping a coin), where each trial has two possible outcomes: success and failure. The possible values of \( r \) range from 0 (no successes in all trials) to \( n \) (success in every trial). Therefore, the sample space for \( r \) consists of the set \( \{0, 1, 2, \ldots, n \} \). The sum of the probabilities of achieving each of these outcomes (from 0 successes to n successes) must be 1 because one of these outcomes must occur in any set of \( n \) trials.
02

Break Down P(r ≥ 1) = 1 - P(0)

The event \( r \geq 1 \) corresponds to having at least one success. The complement of this event (\( r < 1 \)) is having zero successes, i.e., \( r = 0 \). By the complement rule, \( P(r \geq 1) \) equals 1 minus the probability of its complement. Therefore, \( P(r \geq 1) = 1 - P(0) \).
03

Understand P(r ≥ 2) = 1 - P(0) - P(1)

The event \( r \geq 2 \) means having at least 2 successes. The complement of this event is having fewer than 2 successes (i.e., \( r = 0 \) or \( r = 1 \)). By the complement rule, \( P(r \geq 2) \) equals 1 minus the probability of its complement, which includes \( P(0) \) and \( P(1) \). Thus, \( P(r \geq 2) = 1 - P(0) - P(1) \).
04

Generalize to P(r ≥ m) = 1 - P(0) - P(1) - ... - P(m-1)

The event \( r \geq m \) corresponds to having at least \( m \) successes. The complement of this event is having fewer than \( m \) successes, leading to the possible values \( r = 0, 1, \ldots, m-1 \). Thus, by the complement rule, the probability \( P(r \geq m) \) is 1 minus the sum of probabilities of having \( r = 0, 1, \ldots, m-1 \) successes. Therefore, \( P(r \geq m) = 1 - P(0) - P(1) - \cdots - P(m-1) \).

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Key Concepts

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

Sample Space
In the context of binomial distributions, the sample space is a fundamental concept that refers to all possible outcomes of an experiment. Imagine you are flipping a coin several times, seeking heads as your success. If you flip the coin only once, the outcomes can either be heads (a success) or tails (a failure). In binomial terms, these are the values 0 and 1.

Now, let's consider a scenario with more than one coin flip. For a binomial distribution where you flip a coin \( n \) times, each trial can result in a distinct number of successes, ranging from 0 to \( n \). Therefore, the sample space for a binomial random variable, \( r \), is \( \{0, 1, 2, \ldots, n\} \).

The reason the probabilities within this sample space must sum to 1 is because one of these outcomes will certainly occur in any series of \( n \) trials. The certainty of an outcome occurring in probabilistic terms is represented by 1. Thus, for a complete sample space, all probability values add up to affirm that certainty.
Complement Rule
The Complement Rule is a helpful probabilistic concept that allows you to find the probability of an event by considering its opposite or complement. Imagine you want to find the probability that you get at least one success in a given scenario. Instead of directly calculating it, a more efficient method may be at hand by considering the opposite case first.

For example, let's say the probability of a certain event, \( A \), is not known. However, it is often easier to determine the probability of the event not happening, denoted by \( A^c \). The sum of the probability of both must equal 1, i.e., \( P(A) + P(A^c) = 1 \).

Thus, in binomial distributions, if you want to know the probability of at least one success \( P(r \geq 1) \), you can easily compute \( 1 - P(0) \), where \( P(0) \) is the probability of no successes.
Probability
Probability is the measure of how likely an event is to occur, and in the context of the binomial distribution, it calculates the likelihood of a certain number of successes in \( n \) independent trials. Each trial is a separate event where the expected outcome is tracked.

Consider the basic concept of probability as a fraction of favorable outcomes over the total possible outcomes. If a single coin flip has a chance of heads as \( \frac{1}{2} \), then flipping more times increases the number of possible outcomes exponentially, yet probability calculations remain essential in making sense of potential success.

For a binomial random variable described by parameters \( n \) and \( p \), the probability of exactly \( k \) successes, \( P(r = k) \), can be calculated using the formula:
  • \[ \binom{n}{k} p^k (1-p)^{n-k} \]
Where \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) successes out of \( n \) trials.
Independent Trials
In probability and statistics, one of the essential concepts when working with binomial distributions is the idea of independent trials. In simple terms, independent trials mean that the outcome of one trial does not affect the outcome of another. Suppose you're rolling a dice multiple times. Each roll is independent because the result of one roll does not change what happens in subsequent rolls.

For a binomial situation, each coin flip, dice roll, or any repeated action targeting success or failure is treated as an independent trial. When these trials are independent, it simplifies the computation of probabilities, as the outcome of one does not mistake the probabilities of outcomes for another.

With independent trials, you can calculate joint probabilities by simply multiplying the probabilities of each independent event. This independence is what enables the binomial distribution's machinery, making it reliable for predictions when each event is free from the influence of another.

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Most popular questions from this chapter

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