MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Concept explainers

Addition Rule of Probability
It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.
Expected Value
When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known …
Probability Distributions
Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosi…
Basic Probability
The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the prob…
Topic Video
Question
**Question:** Of the cartons produced by a company, 3% have a puncture, 5% have a smashed corner, and 0.4% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner. **Solution:** To find the probability that a randomly selected carton has a puncture or a smashed corner, we need to apply the principle of inclusion-exclusion. The formula to find the probability of the union of two events A and B is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \( P(A) \) is the probability of having a puncture = 3% = 0.03 - \( P(B) \) is the probability of having a smashed corner = 5% = 0.05 - \( P(A \cap B) \) is the probability of having both a puncture and a smashed corner = 0.4% = 0.004 Plug these values into the formula: \[ P(A \cup B) = 0.03 + 0.05 - 0.004 = 0.076 \] Convert the decimal back to a percentage: \[ P(A \cup B) = 7.6\% \] Thus, the probability that a randomly selected carton has a puncture or a smashed corner is **7.6%**.
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Transcribed Image Text:**Question:** Of the cartons produced by a company, 3% have a puncture, 5% have a smashed corner, and 0.4% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner. **Solution:** To find the probability that a randomly selected carton has a puncture or a smashed corner, we need to apply the principle of inclusion-exclusion. The formula to find the probability of the union of two events A and B is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \( P(A) \) is the probability of having a puncture = 3% = 0.03 - \( P(B) \) is the probability of having a smashed corner = 5% = 0.05 - \( P(A \cap B) \) is the probability of having both a puncture and a smashed corner = 0.4% = 0.004 Plug these values into the formula: \[ P(A \cup B) = 0.03 + 0.05 - 0.004 = 0.076 \] Convert the decimal back to a percentage: \[ P(A \cup B) = 7.6\% \] Thus, the probability that a randomly selected carton has a puncture or a smashed corner is **7.6%**.
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