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Chapter 4: Problem 18

The probability that an egg in a retail package is cracked or broken is \(0.025 .\) a. Find the probability that a carton of one dozen eggs contains no eggs that are either cracked or broken. b. Find the probability that a carton of one dozen eggs has (i) at least one that is either cracked or broken; (ii) at least two that are cracked or broken. c. Find the average number of cracked or broken eggs in one dozen cartons.

Short Answer

Expert verified
a) 0.7351, b-i) 0.2649, b-ii) 0.0399, c) 0.3 eggs.

Step by step solution

01

Understanding the Scenario

The task is to find probabilities related to the condition of eggs in a carton of 12, where each egg has a 0.025 probability of being cracked or broken. This probability reflects a binomial distribution scenario.
02

Calculating Probability of No Broken Eggs (Part a)

For part (a), we use the formula for the binomial probability of exactly 0 successes (cracked eggs) in 12 trials (eggs), where the probability of a success (a cracked egg) is 0.025: \[ P(X = 0) = \binom{12}{0} (0.025)^0 (0.975)^{12} \]Calculating this, we find:\[ P(X = 0) = 1 \times 1 \times 0.73509189 \approx 0.7351 \]
03

Calculating Probability of At Least One Cracked Egg (Part b-i)

For part (b-i), we need the probability of at least one cracked egg: \[ P(X \geq 1) = 1 - P(X = 0) \]This equates to:\[ P(X \geq 1) = 1 - 0.7351 = 0.2649 \]
04

Calculating Probability of At Least Two Cracked Eggs (Part b-ii)

For part (b-ii), calculate the probability of at least two cracked eggs by subtracting the probability of 0 or 1 cracked egg from 1: \[ P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \]\[ P(X = 1) = \binom{12}{1} (0.025)^1 (0.975)^{11} \]Computing this gives:\[ P(X = 1) = 12 \times 0.025 \times 0.7513148 \approx 0.225 \P(X \geq 2) = 1 - (0.7351 + 0.225) = 0.0399 \]
05

Calculating the Expected Number of Cracked Eggs (Part c)

For part (c), the expected number of cracked eggs in a dozen is given by the formula for expected value in a binomial distribution: \[ E(X) = np \]where \( n = 12 \) and \( p = 0.025 \). Thus:\[ E(X) = 12 \times 0.025 = 0.3 \]
06

Conclusion

We have calculated all the required probabilities and the expected number of cracked eggs in a dozen carton.

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Key Concepts

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

Probability
Probability is a measure of the likelihood of an event occurring. In the context of our problem, it deals with the chance that an egg in a carton is cracked or broken. To solve this type of problem, we use the binomial distribution because each egg has the same probability of being cracked independently of the others.
In this problem:
  • The probability of an egg being cracked (success) is given as 0.025.
  • The probability of an egg not being cracked (failure) is therefore 1 - 0.025 = 0.975.
Using the binomial formula, we calculate various probabilities by considering these two outcomes. For example, the probability of finding no cracked eggs in a carton uses the binomial probability formula for 0 successes in 12 trials. This calculation involves raising the probability of not being cracked (0.975) to the twelfth power, as all eggs must not be cracked to satisfy the condition. This results in a probability of around 0.7351 for having no cracked eggs in a carton.
Expected Value
Expected value is a concept that provides the average outcome of a random event over many trials. In our problem, we are interested in the average number of cracked eggs in a dozen-carton situation.
For a binomial distribution, the expected value can be calculated using the formula \[ E(X) = np \]where:
  • \( n \) is the number of trials (in this case, 12 eggs).
  • \( p \) is the probability of success (an egg being cracked), which is 0.025.
By substituting these values, we find that: \[ E(X) = 12 \times 0.025 = 0.3 \]So, on average, a carton of 12 eggs is expected to contain 0.3 cracked eggs. Remember, this expected value is a prediction of the average trend over many cartons, not a guarantee for any specific one carton.
Cracked Eggs
Understanding the probability of cracked eggs is key to making informed decisions based on risk鈥攕uch as knowing how likely it is you'll find a cracked egg in your purchase or handling logistics in egg distribution.
When working with cracked eggs in this problem, the binomial model is used to calculate different scenarios:
  • Finding the probability of zero cracked eggs involves the equation \( P(X = 0) \), which we computed to approximately 0.7351.
  • To find at least one cracked egg, calculate the complementary probability: \( P(X \geq 1) = 1 - P(X = 0) \).
  • To find the probability of at least two cracked eggs, you subtract the probabilities of having 0 and 1 cracked eggs from 1. For 1 cracked egg, use \( P(X = 1) \), which account for one egg being cracked out of twelve.
By understanding these probabilities, you can grasp how often cracked eggs might appear and manage expectations when purchasing or selling egg cartons.

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Most popular questions from this chapter

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