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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 are the possible values of \(y\) ? b. What is \(p(2)=P(y=2) ?\) c. What is \(p(3) ?\) (Hint: There are two different outcomes that result in \(y=3\).)

Short Answer

Expert verified
a. The possible values for y are any integer 2 or greater. b. The value of P(y=2), the probability that exactly two batteries need to be tested to find two acceptable ones, is 0.64. c. The value of P(y=3), the probability that three batteries need to be tested to find two acceptable ones, is 0.288.

Step by step solution

01

Determine possible values for y

Since two acceptable batteries are needed, the minimum number of batteries that must be tested is two. So the smallest possible value of y is 2. Since there is a chance that a battery could be unacceptable, there is no upper limit to the possible values of y. In other words, y could be any positive integer 2 or greater, i.e. y ∈ {2, 3, 4, 5, ...}.
02

Calculate P(y=2)

The value P(y=2) will represent the scenario where both tested batteries are acceptable. This would be a simple case of multiplying the independent probabilities together. Since the probability of a battery being acceptable is 0.8, the equation is \(P(y=2) = 0.8 × 0.8 = 0.64\). So, the probability that exactly 2 batteries are required to be tested to find two acceptable batteries is \(0.64\).
03

Calculate P(y=3)

Calculating P(y=3) is a bit more complex since we have to consider two possibilities: the first two batteries tested could both be unacceptable and the third could be acceptable, or the first battery could be unacceptable and the next two could be acceptable. For the first case, we have probability \(P = 0.2 × 0.2 × 0.8 = 0.032\), for the second case we have probability \(P = 2 × (0.2 × 0.8 × 0.8 = 0.128)\). We multiply the second case by 2 as this event can happen in two ways (either the first or the second battery tested is unacceptable). Adding these two possibilities together, we get \(P(y=3) = 0.032 + 0.128 × 2 = 0.288\).

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Key Concepts

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

Independent Events
Independent events are outcomes where the occurrence of one event does not affect the probability of the other. In the context of the battery testing exercise, each battery is tested individually, and whether or not a battery is acceptable does not influence the result of the next battery test. This is why the probabilities can be directly multiplied.

For instance, consider the probability of both batteries being acceptable if each has a probability of 0.8 of being acceptable. Since these events are independent, we calculate the combined probability as 0.8 multiplied by 0.8, which equals 0.64. This calculation confirms that both batteries being acceptable is a separate event not influenced by each other, showcasing the principle of independent events.

This principle simplifies the process of analyzing sequences of independent tests, such as the battery example, ensuring each event's outcome is purely based on its own probability.
Uniform Distribution
Uniform distribution represents a situation where all outcomes have equal probability of occurrence. However, in the given exercise, the probability distribution of finding two acceptable batteries is not uniform.

Instead, the task utilizes a probability distribution based on independent events. This means each trial for a battery being acceptable is consistent but not equally probable across all possible values of batteries tested. In a truly uniform distribution scenario, every result would have an equal chance, which is not the case here.

The distribution in this case tends to favor outcomes where fewer batteries are tested successfully due to the higher probability of a battery being acceptable (0.8), contrasting the nature of a uniform distribution.
Expected Value
The expected value is a fundamental concept in probability which represents the average or mean value that would result from a significant number of trials. It shows the long-term average or expected outcome when randomness is involved.

For this battery testing problem, calculating the expected value entails determining how many batteries, on average, will need to be tested before two acceptable ones are found. Each test result is weighted by its probability and summed to get the expected number of tests.

While the solution does not compute this directly, understanding the concept helps anticipate the average number of tries needed based on given probabilities. Since 80% of batteries are acceptable, fewer tests are generally needed, suggesting a smaller expected value. This highlights the utility of expected value in predicting outcomes over numerous trials without performing each trial individually.

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Most popular questions from this chapter

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