Question

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 \(80 \%\) of all batteries have acceptable voltages, and let \(y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(y=2)\) ? b. What is \(p(3) ?\) (Hint: There are two different outeomes that result in \(y=3 .\) ) c. In order 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

The probabilities of the number of trials needed to find two acceptable batteries are: p(2) = 0.64, p(3) = 0.256, and p(5) = 0.02048. The general formula for p(y) is: \(P(y)={y-1 \choose 2}*(0.8)^2*(0.2)^{y-2}\).

Step-by-step solution

Calculate probability for \(y=2\)

For \(y=2\), the only possible scenario is that both tested batteries are acceptable. Hence, \(P(y=2)\) equals the probability of success (A) on the first and the second trials: \(P(y=2)= (0.8)^2 = 0.64\).

Calculate probability for \(y=3\)

For \(y=3\), there are two possible scenarios: the first battery is unacceptable and the next two are acceptable (UAA), or the first two batteries are acceptable but the third is unacceptable (AAU). The probability of these outcomes are, respectively, \(0.2(0.8)^2\) and \((0.8)^2*0.2\). So \(P(y=3)=2* (0.8)^2 *0.2 = 0.256\).

Calculate probability for \(y=5\)

For \(y=5\), the fifth battery must be the second acceptable one. Thus, all scenarios contain exactly two A's and three U's. The possible outcomes are UUUAA, UUAUA, UAUUA, AUUUA. Their respective probabilities are \((0.8)^2 (0.2)^3\), hence \(P(y=5)=4*((0.8)^2 (0.2)^3) = 0.02048\).

Generalize the formula for \(p(y)\)

The pattern in the solutions suggests that \(P(y)\) is determined by the number of ways to distribute two A's among \(y-1\) positions, times the probability of that outcome. This suggests the use of the binomial coefficient. Substituting \(n=y-1\) and \(k=2\) (the number of successes) in the binomial formula we get \(P(y)={y-1 \choose 2}*(0.8)^2*(0.2)^{y-2}\).

Key concepts

Probability Theory

At the heart of understanding events like the selection of batteries lies probability theory. It is the branch of mathematics concerned with the analysis of random phenomena. The probability of an event is a measure of the likelihood that the event will occur, and it ranges from 0 (impossible event) to 1 (certain event).

When dealing with independent events, like the selection of batteries in our exercise, the probability of multiple events occurring is the product of their individual probabilities. For example, in the case of the first two batteries being acceptable (event A), the probability of this occurrence, assuming independence, is calculated by multiplying the probability of one A event by itself, or \(P(y=2) = P(A) \times P(A)\).

However, when an event can occur in multiple ways, we sum the probabilities of each different way. This is why for \(P(y=3)\), the event can occur as UAA or AAU, and we add these probabilities to find the total.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and the counting of various configurations. It plays a crucial role in calculating probabilities when we need to determine the number of ways an event can occur.

In our battery selection problem, when we calculate \(P(y=5)\), we use combinatorics to find all possible sequences of acceptable (A) and unacceptable (U) batteries that result in the fifth battery being the second acceptable one. The term 'combinatorics' in this context refers to the various combinations of two As and three Us. For instance, the sequence UAUUA is one such combination.

To calculate the total probability for \(y=5\), we determine the number of combinations of these two A’s within the five attempts, while taking into account that the fifth attempt must be A. We then multiply this by the probability of each sequence occurring, which involves raising the probability of A and U to the powers corresponding to how many times they occur in each sequence.

Statistical Analysis

Statistical analysis helps us interpret and draw conclusions from data by using various techniques. In the case of our probability exercise, we can use statistical analysis to predict the behavior of the battery selection process.

By analyzing the probabilities we have calculated through the combinatoric approach, for instance, the general formula for \(p(y)\), we can determine the likelihood of certain outcomes over a large number of trials. This is directly related to statistical concepts such as expected value and variance which tell us about the average outcome we might expect and how much the results might spread out or vary.

Frequently, statistical analysis in such contexts also involves creating models to fit the data observed; in this case, we’ve identified the binomial probability distribution as an appropriate model, which then allows us to derive general formulas to predict future events. The outcomes from parts (a) to (c) of the problem illustrate the kind of patterns that we often look for during the analysis to develop such predictive models.