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Chapter 3: Problem 57

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?

Short Answer

Expert verified
The probability that at least nine flashlights work is 0.569, assuming independence and equal probability of acceptable batteries.

Step by step solution

01

Define the Random Variables

Let the random variable \( X \) be the number of flashlights that work. Since a flashlight works if both batteries have acceptable voltages, and there is a 90% chance for a single battery to be acceptable, we calculate the probability that a flashlight works. Assuming independence, the probability a flashlight works is \((0.9) \times (0.9) = 0.81\). Thus, \( X \) follows a binomial distribution: \( X \sim \text{Binomial}(10, 0.81) \).
02

Calculate the Desired Probability

We are interested in the probability that at least 9 out of 10 flashlights work, which is \( P(X \geq 9) \). This can be calculated using the cumulative probability: \( P(X = 9) + P(X = 10) \).
03

Calculate P(X=9)

For \( P(X = 9) \), use the binomial probability formula: \[P(X = 9) = \binom{10}{9} (0.81)^9 (1 - 0.81)^1 \]Calculating, we find:\[ \binom{10}{9} = 10 \]\[ (0.81)^9 \approx 0.194 \]\[ (0.19) = 0.19 \]Thus \[ P(X = 9) = 10 \times 0.194 \times 0.19 = 0.368 \]
04

Calculate P(X=10)

For \( P(X = 10) \), use the binomial probability formula: \[P(X = 10) = \binom{10}{10} (0.81)^{10} (1 - 0.81)^0 \]Calculating, we find:\[ \binom{10}{10} = 1 \]\[ (0.81)^{10} \approx 0.201 \]Thus \[ P(X = 10) = 1 \times 0.201 \times 1 = 0.201 \]
05

Sum Probabilities for Final Solution

Add \( P(X = 9) \) and \( P(X = 10) \) to find \( P(X \geq 9) \):\[ P(X \geq 9) = P(X = 9) + P(X = 10) = 0.368 + 0.201 = 0.569 \]
06

State Assumptions Made

Assumptions include that the acceptance of each battery is independent, each battery has the same probability of being acceptable, and each flashlight requires exactly two batteries. The performance of each flashlight was assumed to be independent.

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Key Concepts

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

Probability Calculation
Probability calculation is an essential concept in understanding binomial distribution problems. In this exercise, we're determining the likelihood that a particular number of flashlights will work based on certain conditions. First, identify the event you are interested in鈥攖he number of flashlights with acceptable batteries. Then, calculate the probability for each scenario.

Given that each battery has a 90% acceptance probability, a single flashlight with two batteries has a probability of both being acceptable, calculated as follows:
  • Single battery acceptance probability = 0.9
  • Flashlight working probability = 0.9 (for the first battery) x 0.9 (for the second battery) = 0.81
This implies that 81% of the flashlights should work based on the probabilities provided. Calculating probabilities for individual numbers of functioning flashlights requires using the binomial probability formula, which defines the probability of a certain number of successes across many trials.
Independent Events
Independent events are a key assumption in this problem. Batteries' acceptance rates must not influence one another for independence to hold. Essentially, whether one battery is acceptable does not affect the other battery's likelihood of being acceptable. Each battery's acceptability is treated as an independent event.

Understanding independence:
  • If event A and event B are independent, the probability of both occurring is the product of their probabilities.
  • For the flashlight, this means the probability that both batteries are acceptable is calculated by multiplying their separate probabilities, as done earlier.
  • In this case: 0.9 x 0.9 = 0.81.
Such independence simplifies the calculation of probabilities and is a foundational concept in calculating probabilities for combined events.
Cumulative Probability
Cumulative probability is vital for answering the exercise's main question: "What is the probability that at least nine flashlights will work?"

When calculating cumulative probability, we're summing up the probabilities of several events. In this exercise, it involves combining the probability of exactly nine flashlights working with the probability of all ten working. These events are mutually exclusive, meaning they cannot happen simultaneously, yet both contribute to the total probability of at least nine working. Thus, we calculate:
  • Probability of 9 flashlights working: \[ P(X = 9) = \binom{10}{9} (0.81)^9 (0.19)^1 \]
  • Probability of 10 flashlights working: \[ P(X = 10) = \binom{10}{10} (0.81)^{10} (0.19)^0 \]
Once these are determined, adding them gives the final solution:
  • \[ P(X \geq 9) = P(X = 9) + P(X = 10) \]
The cumulative probability succinctly represents the probability that the number of working flashlights is within or exceeds a certain range.
Random Variables
Random variables are concepts that assign numerical values to the outcomes of random phenomena. In this particular problem, we introduce a random variable \( X \), which denotes the number of flashlights functioning based on the given battery conditions.

Understanding \( X \) as a random variable is crucial:
  • \( X \) follows a binomial distribution: \( X \sim \text{Binomial}(n, p) \), where \( n = 10 \) flashlights, and \( p = 0.81 \) is the probability of each flashlight working
  • The binomial distribution here models the number of successes (working flashlights) in a fixed number of trials (10 flashlights)
This type of distribution is applicable when each trial is independent, and there are two possible outcomes: success or failure鈥攊n this case, a flashlight either works or it doesn't.

Using random variables in this context helps organize our probability calculations, turning complex scenarios into structured probabilistic models.

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Most popular questions from this chapter

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