Problem 22
Question
One light bulb is selected at random from a lot of 120 light bulbs, of which \(5 \%\) are defective. What is the probability that the light bulb selected is defective?
Step-by-Step Solution
Verified Answer
The probability of selecting a defective light bulb from the lot is \(\frac{1}{20}\) or 0.05 (5%).
1Step 1: Find the total number of possible outcomes
The total number of possible outcomes is the total number of light bulbs in the lot, which is 120.
2Step 2: Find the number of favorable outcomes
Since 5% of the light bulbs are defective, we can find the number of defective bulbs by multiplying the total number of bulbs by the percentage of defective bulbs:
Number of defective bulbs = \(0.05 \times 120 = 6\)
3Step 3: Calculate the probability of selecting a defective bulb
Now, we can use the definition of probability to find the probability of selecting a defective bulb:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = \(6 / 120\)
4Step 4: Simplify the probability
We can simplify the fraction 6/120 by dividing both the numerator and the denominator by their greatest common divisor (GCD), which is 6:
Probability = \(6 / 120 = 1 / 20\)
So, the probability that a randomly selected light bulb from the lot is defective is 1/20, or 0.05 (5%).
Key Concepts
Defective Items ProbabilityProbability CalculationBasic Statistics
Defective Items Probability
Understanding the probability of selecting defective items is crucial in quality control and inventory management. In our example, we're examining a lot of 120 light bulbs, out of which 5% are known to be defective. To find the probability of picking a defective light bulb randomly, we first need to quantify what 'defective' means in this context.
It's given that 5% of the total amount represents the defective light bulbs. This implicitly tells us that for every 100 light bulbs, 5 are defective. Extrapolating this to 120 light bulbs, simple arithmetic (\(0.05 \times 120\)) gives us the number of defective bulbs, which is 6. Thus, if we were to select one bulb at random, the 'favorable outcome'—the event we're concerned about—is choosing one of these 6 defective bulbs. This forms the basis for our probability calculation.
It's given that 5% of the total amount represents the defective light bulbs. This implicitly tells us that for every 100 light bulbs, 5 are defective. Extrapolating this to 120 light bulbs, simple arithmetic (\(0.05 \times 120\)) gives us the number of defective bulbs, which is 6. Thus, if we were to select one bulb at random, the 'favorable outcome'—the event we're concerned about—is choosing one of these 6 defective bulbs. This forms the basis for our probability calculation.
Probability Calculation
The essence of probability calculation lies in this fundamental idea: probability is the ratio of the number of favorable outcomes to the total number of possible outcomes. More formally, it is expressed as \[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
In our scenario, the favorable outcomes are the 6 defective light bulbs, and the total number of outcomes is all 120 bulbs in the lot. Applying the formula gives us \[ P(\text{defective}) = \frac{6}{120} \]. This fraction can be simplified to \[ \frac{1}{20} \], which reflects the probability of a single bulb being defective. The significance of simplifying the fraction is not just for aesthetics; it makes the probability easier to understand and communicate — critical aspects when explaining concepts to others.
In our scenario, the favorable outcomes are the 6 defective light bulbs, and the total number of outcomes is all 120 bulbs in the lot. Applying the formula gives us \[ P(\text{defective}) = \frac{6}{120} \]. This fraction can be simplified to \[ \frac{1}{20} \], which reflects the probability of a single bulb being defective. The significance of simplifying the fraction is not just for aesthetics; it makes the probability easier to understand and communicate — critical aspects when explaining concepts to others.
Basic Statistics
Basic statistics provides us with tools and methods to systematically analyze data and draw conclusions. In the realm of probability, it allows us to understand the likelihood of certain events occurring within a set of given conditions. Our investigation into the defective light bulbs serves as an excellent example of applying basic statistical methods to a real-world problem.
The process of calculating the probability of finding a defective light bulb incorporates several fundamental statistical concepts: percentage calculation, multiplication for determining quantitative relationships, and ratio simplification to calculate the final probability. These procedures exemplify how basic statistics can offer practical, useful insights, which in this case, can aid in making informed decisions about the light bulb lot, such as determining the quality of the batch or whether additional quality control measures might be warranted based on the probability of defectives.
The process of calculating the probability of finding a defective light bulb incorporates several fundamental statistical concepts: percentage calculation, multiplication for determining quantitative relationships, and ratio simplification to calculate the final probability. These procedures exemplify how basic statistics can offer practical, useful insights, which in this case, can aid in making informed decisions about the light bulb lot, such as determining the quality of the batch or whether additional quality control measures might be warranted based on the probability of defectives.
Other exercises in this chapter
Problem 21
In a poll conducted among 200 active investors, it was found that 120 use discount brokers, 126 use fullservice brokers, and 64 use both discount and full-servi
View solution Problem 21
Find the smallest possible set (i.e., the set with the least number of elements) that contains the given sets as subsets. $$ \\{1,2\\},\\{1,3,4\\},\\{4,6,8,10\\
View solution Problem 22
Evaluate the given expression. $$ C(n, n-2) $$
View solution Problem 22
A warranty identification number for a certain product consists of a letter of the alphabet followed by a five-digit number. How many possible identification nu
View solution