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Chapter 3: Problem 17

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is \(P(Y=2)\) ? b. What is \(p(3)\) ? c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

Short Answer

Expert verified
a. 0.81 b. 0.081 c. Fifth must be acceptable; four sequences lead to 0.00324 Formula: \( P(Y = y) = \binom{y-1}{2} \times 0.1^{y-2} \times 0.9^{2} \).

Step by step solution

01

Understand Probability of Acceptable Batteries

The probability that a battery is acceptable is given as 0.9. Therefore, the probability that a battery is unacceptable is 0.1.
02

Calculate p(2) for Y=2

The event that exactly 2 batteries are required to find two acceptable ones occurs when both batteries are acceptable. Therefore, \( P(Y = 2) = 0.9 \times 0.9 = 0.81 \).
03

Calculate p(3) for Y=3

To have \( Y = 3 \), the first battery must be unacceptable, and the next two batteries must be acceptable. Thus, \( P(Y = 3) = 0.1 \times 0.9 \times 0.9 = 0.081 \).
04

Analyze conditions for Y=5

For \( Y = 5 \), the first four batteries must include exactly three unacceptable ones and the fifth one must be acceptable. Possible sequences are: UUUA, UUAU, UAUA, AUUA where U denotes unacceptable, A denotes acceptable, and the last battery (5th) is always acceptable.
05

Compute p(5)

For each sequence where the fourth battery is unacceptable, the probability is \( 0.1^3 \times 0.9^2 \). Counting all such sequences, there are 4 possibilities, so \( P(Y = 5) = 4 \times 0.1^3 \times 0.9^2 = 0.00324 \).
06

Develop a General Formula for p(y)

The general formula for \( P(Y = y) \) requires \( y-1 \) total batteries with \( y-2 \) unacceptable and 2 being acceptable. Thus, \( P(Y = y) = \binom{y-1}{2} \times 0.1^{y-2} \times 0.9^{2} \), where \( \binom{y-1}{2} \) gives the number of ways to choose which \( y-2 \) batteries are unacceptable out of the first \( y-1 \).

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Key Concepts

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

Probability
Probability is a concept that measures the likelihood of an event happening. It ranges from 0 to 1, where 0 indicates impossibility and 1 means certainty. We use probability to analyze uncertain situations, such as when deciding if a battery will have acceptable voltage or not.

In the case of the problem we're dealing with, the probability of a battery being acceptable is 0.9. This means there is a 90% chance that any given battery will be good. Conversely, there is a 0.1 (or 10%) probability that it will be unacceptable. Calculating probabilities usually involves multiplying the probabilities of independent events, which we'll talk about next.
Independent Events
Independent events are when the outcome of one event does not affect the outcome of another. In simple words, what happens with one event doesn't change what happens with the other. This concept is crucial when calculating probabilities in a situation where multiple events occur.

In our battery example, each battery is independent of others, meaning whether the first battery is acceptable does not affect whether the second battery is acceptable. Because of this independence, we can calculate the overall probability by multiplying individual probabilities, like how we calculated for the event where both batteries must be acceptable for the flashlight by multiplying their probabilities: 0.9 x 0.9.
Random Variable
A random variable is a numerical representation of outcomes from a random process. It helps translate the outcomes into understandable values.In this exercise, the random variable, denoted by \(Y\), represents the number of batteries we have to test to find two acceptable ones.

This random variable takes different possible values such as 2, 3, 5, etc., depending on how many batteries need to be tested before finding two that are acceptable. It basically counts the trials until we reach the desired outcome, and we can use this count to find the probability of each scenario occurring.
Binomial Coefficients
Binomial coefficients help us calculate how many ways we can arrange a certain number of items. They are part of what's called binomial distribution, which is useful in our battery testing problem.

For instance, to determine how many ways to arrange unacceptable batteries before a specific number of acceptable ones, we use binomial coefficients. In a formula, represented as \( \binom{n}{k} \), where \(n\) is the total number of trials and \(k\) is the number of successful outcomes (or arrangements in this context). For example, \( \binom{y-1}{y-2} \) helps to determine how many ways \(y-2\) unacceptable batteries can be placed among the first \(y-1\) trials.

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Most popular questions from this chapter

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).

The number of pumps in use at both a six-pump station and a four-pump station will be determined. Give the possible values for each of the following random variables: a. \(T=\) the total number of pumps in use b. \(X=\) the difference between the numbers in use at stations 1 and 2 c. \(U=\) the maximum number of pumps in use at either station d. \(Z=\) the number of stations having exactly two pumps in use

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?

Suppose that \(p=P\) (male birth \()=.5\). A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled. a. What is the probability that the family has \(x\) male children? b. What is the probability that the family has four children? c. What is the probability that the family has at most four children? d. How many male children would you expect this family to have? How many children would you expect this family to have?

A particular type of tennis racket comes in a midsize version and an oversize version. Sixty percent of all customers at a certain store want the oversize version. a. Among ten randomly selected customers who want this type of racket, what is the probability that at least six want the oversize version? b. Among ten randomly selected customers, what is the probability that the number who want the oversize version is within 1 standard deviation of the mean value? c. The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?
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