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Chapter 3: Problem 63

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

Short Answer

Expert verified
Parts (a) and (b) show symmetry, so tables only need \(p \leq 0.5\).

Step by step solution

01

Understand the Problem

We are given two main problems to prove about the binomial probabilities and cumulative distribution functions with a hint that relates success and failure probabilities. Our task is to prove part (a) regarding individual binomial probabilities and part (b) regarding cumulative binomial probabilities, and then interpret what they imply about the probability table values.
02

Recall the Binomial Probability Formula

The binomial probability formula is given by \[b(x; n, p) = \binom{n}{x} p^x (1-p)^{n-x}\]where \(b(x; n, p)\) is the probability of getting \(x\) successes in \(n\) trials.
03

Prove Part (a)

To prove \(b(x ; n, 1-p) = b(n-x ; n, p)\), substitute in the probability formula:\[b(x ; n, 1-p) = \binom{n}{x} (1-p)^x p^{n-x}\]and\[b(n-x ; n, p) = \binom{n}{n-x} p^{n-x} (1-p)^x\].Since \(\binom{n}{x} = \binom{n}{n-x}\), both expressions are equal.
04

Interpret Part (a)

Part (a) shows that the distribution is symmetric with respect to \(p\) and \(1-p\), supporting that a binomial distribution with success probability \(1-p\) can be expressed as a failure probability \(p\).
05

Prove Part (b)

For \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\), note that:\[B(x ; n, 1-p) = \sum_{k=0}^{x} b(k; n, 1-p)\]\[1-B(n-x-1 ; n, p) = \sum_{k=n-x}^{n} b(k; n, p)\].Given the hint, at most \(x\) successes is equivalent to at least \((n-x)\) failures, thus the two sums cover complementary events of the entire probability space.
06

Interpret Part (b)

Part (b) implies a duality in cumulative binomial distributions similar to part (a), using the symmetry in tail sums of the cumulative probability function.
07

Implication for Probability Tables

Parts (a) and (b) suggest that due to symmetry in \(p\) and \(1-p\), it is unnecessary to include \(p > 0.5\) in probability tables like Appendix Table A.1, as values for \(p > 0.5\) can be derived from corresponding values of \(1-p\).

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Key Concepts

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

Probability Distribution
When discussing binomial probability, it's all about understanding **Probability Distribution**. In the binomial distribution, we are looking at the likelihood of a certain number of successes across multiple trials. This concept is represented by the formula: \[b(x; n, p) = \binom{n}{x} p^x (1-p)^{n-x}\]Here,
  • \(n\) is the total number of trials,
  • \(x\) is the number of desired successes,
  • \(p\) is the probability of success in each trial.
This formula helps in calculating the specific probability for any given number of successes \(x\). The distribution is discrete, as it's about the possible outcomes of a finite number of trials. When using binomial probability distributions, you must remember that they only apply when the trials are independent and the probability of success remains constant.
Cumulative Distribution Function
The **Cumulative Distribution Function (CDF)** in binomial probability displays the total probability of getting a specified number of successes or fewer. In simpler terms, it sums up the probabilities for all outcomes up to a certain point. It's given by:\[B(x ; n, p) = \sum_{k=0}^{x} b(k; n, p)\]This function is especially useful because it helps to answer questions like, "What is the probability of having at most \(x\) successes?" For instance, if you needed to know the probability of having no more than 3 successes in 10 trials with a success probability of 0.6, the CDF could be employed to sum probabilities from 0 to 3. Notably, in part (b) of the exercise, we explored that at most \(x\) successes can imply at least \((n-x)\) failures. This hint is the heart of how the CDF works from different angles.
Symmetry in Probability
**Symmetry in Probability** comes into play significantly with the binomial distribution. This symmetry is nicely illustrated in the exercise parts (a) and (b) proved above. It points out that a binomial distribution with probability \(p\) is essentially mirrored when you consider a probability \(1-p\). In simple terms, this symmetry implies:
  • The probability of exactly \(x\) successes when \(p = 1-p'\) mirrors the probability of \(n-x\) failures with \(p' = p\).
  • The cumulative probability up to a certain number of successes with probability \(p\) corresponds to the complementary cumulative tail probability with \(1-p\).
  • This implies you don't need to use tables with \(p > 0.5\), as values can be derived from those for \(p < 0.5\).
This symmetric property bears great practical utility, as it reduces complexity in calculations and resources needed for probability tables.

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Most popular questions from this chapter

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(.1\) mile. Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres. \(]\)

Consider a communication source that transmits packets containing digitized speech. After each transmission, the receiver sends a message indicating whether the transmission was successful or unsuccessful. If a transmission is unsuccessful, the packet is re-sent. Suppose a voice packet can be transmitted a maximum of 10 times. Assuming that the results of successive transmissions are independent of one another and that the probability of any particular transmission being successful is \(p\), determine the probability mass function of the rv \(X=\) the number of times a packet is transmitted. Then obtain an expression for the expected number of times a packet is transmitted.

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An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.
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