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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

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is $$ \begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline p(y) & .60 & .25 & .10 & .05 \end{array} $$ a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let \(X\) be the number among the first 6 examined that have a defective compressor. Compute the following: a. \(P(X=5)\) b. \(P(X \leq 4)\) c. The probability that \(X\) exceeds its mean value by more than 1 standard deviation. d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If \(X\) is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) \(P(X \leq 5)\) than to use the hypergeometric pmf.

For each random variable defined here, describe the set of possible values for the variable, and state whether the variable is discrete. a. \(X=\) the number of unbroken eggs in a randomly chosen standard egg carton b. \(Y=\) the number of students on a class list for a particular course who are absent on the first day of classes c. \(U=\) the number of times a duffer has to swing at a golf ball before hitting it d. \(X=\) the length of a randomly selected rattlesnake e. \(Z=\) the amount of royalties earned from the sale of a first edition of 10,000 textbooks f. \(Y=\) the pH of a randomly chosen soil sample g. \(X=\) the tension (psi) at which a randomly selected tennis racket has been strung h. \(X=\) the total number of coin tosses required for three individuals to obtain a match (HHH or TTT)

Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?

Individuals \(A\) and \(B\) begin to play a sequence of chess games. Let \(S=\\{\) A wins a game \(\\}\), and suppose that outcomes of successive games are independent with \(P(S)=p\) and \(P(F)=1-p\) (they never draw). They will play until one of them wins ten games. Let \(X=\) the number of games played (with possible values \(10,11, \ldots, 19\) ). a. For \(x=10,11, \ldots, 19\), obtain an expression for \(p(x)=\) \(P(X=x)\). b. If a draw is possible, with \(p=P(S), q=P(F), 1-p-\) \(q=P(\) draw \()\), what are the possible values of \(X\) ? What is \(P(20 \leq X)\) ? [Hint: \(P(20 \leq X)=1-P(X<20)\).]
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