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QuaLrr ConTRoL It is estimated that \(0.80 \%\) of a large consignment of eggs in a certain supermarket is broken. a. What is the probability that a customer who randomly selects a dozen of these cggs receives at least one broken egg? b. What is the probability that a customer who selects these eggs at random will have to check three cartons before finding a carton without any broken eggs? (Each carton contains a dozen eggs.)

Short Answer

Expert verified
a. The probability that a customer who randomly selects a dozen of these eggs receives at least one broken egg is approximately 0.0935. b. The probability that a customer who selects these eggs at random will have to check three cartons before finding a carton without any broken eggs is approximately 0.0773.

Step by step solution

01

a. Calculating the probability of at least one broken egg

First, determine the probability of finding an unbroken egg. Let p be the probability of selecting a broken egg, which is 0.008. So, the probability of selecting an unbroken egg (q) is: q = 1 - p = 1 - 0.008 = 0.992 Now, to find the probability of selecting a carton of a dozen eggs without any broken eggs, we will use the binomial distribution formula: \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) where: n = 12 (number of eggs in a carton) k = 0 (since there are no broken eggs) p = 0.008 q = 0.992 \(P(X=0) = \binom{12}{0} (0.008)^0 (0.992)^{12-0}\) \(P(X=0) = (0.992)^{12}\) Now, to find the probability of selecting at least one broken egg, we subtract the probability of selecting a carton without any broken eggs from 1: Prob(at least one broken egg) = 1 - P(X=0) Calculate the probability: Prob(at least one broken egg) = 1 - (0.992)^{12} ≈ 0.0935
02

b. Probability of finding a carton without any broken eggs in three attempts

To find the probability of a customer checking three cartons before finding a carton without any broken eggs, we will use the geometric distribution: \(P(Y=k) = (1-p)^{k-1}p\) In this case, p is the probability of selecting a carton without any broken eggs which we found in part a. So, the probability of finding a carton without broken eggs in three attempts is: \(P(Y=3) = (1-p)^{3-1}p\) \(P(Y=3) = (1-0.0935)^{3-1}(0.0935)\) Calculate the probability: \(P(Y=3) = (0.9065)^2(0.0935) ≈ 0.0773\) So, the probability that a customer who selects these eggs at random will have to check three cartons before finding a carton without any broken eggs is approximately 0.0773.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states and where each trial is identical. In the context of our egg carton example, the two states are finding a broken egg versus finding an unbroken egg each time an egg is picked.

When applying the binomial distribution, we use the formula \(P(X=k) = \binom{n}{k} p^k q^{n-k}\) to calculate the probability of having exactly \(k\) successes in \(n\) independent trials. Here, success could be finding a broken egg (with a probability \(p\)) and failure being an unbroken egg (with a probability \(q = 1 - p\)).

One common misconception is that the binomial distribution is only used when there are 50-50 chances. However, the only requirement is that the trials are independent and the probability of success is consistent across trials. The given exercise required calculating the probability of receiving at least one broken egg in a carton, which was solved by finding the probability of no broken eggs (using \(k = 0\)) and subtracting that from 1.
Geometric Distribution
Unlike the binomial distribution, the geometric distribution deals with the number of trials needed for the first success. It is used when we are interested in knowing how many trials it will take to achieve a success, given that each trial is independent and the probability of success is the same in each trial.

The geometric distribution is described by the formula \(P(Y=k) = (1-p)^{k-1}p\), where \(p\) is the probability of success on any given trial, and \(k\) is the number of trials until the first success. It's important to note that the geometric distribution gives us the probability that the \(k^{th}\) trial is the first successful one, after \(k-1\) failures.

In the egg carton example, the 'success' was defined as finding a carton without any broken eggs. This distribution was used to calculate the probability that on the third carton chosen, for the first time, a customer finds no broken eggs. It is a useful distribution for modeling scenarios where you keep trying until you succeed, and each attempt is independent of the others.
Probability Calculations
Probability calculations are essential in determining how likely it is that an event will occur. These calculations rely on an understanding of the fundamental principles of probability such as outcomes, events, independence, and the structure of various probability distributions.

When calculating the probability of an event, it's crucial to be clear on definitions—like defining a success, determining independence of events, and understanding whether an event is discrete or continuous. Simple probability calculations may involve direct computation from a probability mass function, as with the binomial distribution, or might be based on sequences of trials, as with the geometric distribution.

To enhance understanding, it could be helpful for students to practice converting word problems into probability notation and to work through the logic of why a specific distribution applies to a given situation. In the context of the egg carton exercise, understanding that the event 'selecting a carton without any broken eggs' can be a success or failure—and can be modeled by appropriate distributions—is key to solving the problems presented.

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