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.
The Correct Answer and Explanation is:
Given:
- Probability that a single egg is cracked or broken, p=0.025p = 0.025
- Number of eggs in a dozen = 12
- This is a binomial distribution problem with parameters:
- n=12n = 12
- p=0.025p = 0.025
- q=1−p=0.975q = 1 – p = 0.975
Let X∼Binomial(n=12,p=0.025)X \sim \text{Binomial}(n=12, p=0.025), where XX is the number of cracked or broken eggs in a carton.
a. Probability that a carton has no cracked or broken eggs:
We find P(X=0)P(X = 0): P(X=0)=(120)⋅(0.025)0⋅(0.975)12=(0.975)12≈0.7361P(X = 0) = \binom{12}{0} \cdot (0.025)^0 \cdot (0.975)^{12} = (0.975)^{12} \approx 0.7361
b.
(i) Probability of at least one cracked or broken egg:
P(X≥1)=1−P(X=0)=1−0.7361=0.2639P(X \geq 1) = 1 – P(X = 0) = 1 – 0.7361 = 0.2639
(ii) Probability of at least two cracked or broken eggs:
P(X≥2)=1−P(X=0)−P(X=1)P(X \geq 2) = 1 – P(X = 0) – P(X = 1)
First compute P(X=1)P(X = 1): P(X=1)=(121)⋅(0.025)1⋅(0.975)11=12⋅0.025⋅(0.975)11≈12⋅0.025⋅0.7462≈0.2240P(X = 1) = \binom{12}{1} \cdot (0.025)^1 \cdot (0.975)^{11} = 12 \cdot 0.025 \cdot (0.975)^{11} \approx 12 \cdot 0.025 \cdot 0.7462 \approx 0.2240 P(X≥2)=1−0.7361−0.2240=0.0399P(X \geq 2) = 1 – 0.7361 – 0.2240 = 0.0399
c. Average number of cracked or broken eggs in a carton:
E[X]=n⋅p=12⋅0.025=0.3E[X] = n \cdot p = 12 \cdot 0.025 = 0.3
Explanation
This problem involves calculating probabilities related to the quality of eggs in a retail carton, modeled using the binomial distribution. In this case, each egg independently has a 2.5% chance of being cracked or broken. The binomial distribution is ideal here because we are dealing with a fixed number of independent trials (12 eggs per carton), with each trial resulting in one of two outcomes (cracked or not).
For part (a), we calculate the probability that none of the eggs in the carton is cracked, which is P(X=0)P(X = 0). This is done by raising the probability of an egg being intact (97.5%) to the 12th power, since all eggs must be undamaged.
Part (b) explores defective cartons. For at least one cracked egg, we use the complement rule: subtract the probability of zero cracked eggs from 1. This gives a roughly 26.4% chance of finding at least one cracked egg in a carton. For at least two cracked eggs, we subtract the probabilities of 0 and 1 cracked egg from 1. This probability is lower, about 4%, reflecting the rarity of multiple defects.
Part (c) focuses on the expected number of cracked eggs in a carton. Since the mean of a binomial distribution is given by npnp, we multiply the number of eggs by the defect probability to get 0.3. This suggests that across many cartons, the average number of cracked eggs per carton is 0.3, or about 1 cracked egg every 3 to 4 cartons.
Understanding these probabilities is valuable for quality control in retail food supply chains, helping to anticipate defect rates and improve packaging processes.
