farmer noticed that some of the eggs laid by his hens had double yolks. He stimated the probability of this happening to be 0.05 . Eggs are packed in boxes of 12 . Find the probability that in a box, the number of eggs with double yolks will be exactly one. A customer bought three boxes. Find the probability that only 2 of the boxes contained exactly 1 egg with double yolk. The farmer delivered 10 boxes to a local shop. Using a suitable approximation, find the probability that the delivery contained at least 9 eggs with double yolks. (6 marks) The weight of an individual egg can be modelled by a normal distribution with mean 65 g and standard deviation 2.4 g . Find the probability that a randomly chosen egg weighs more than 68 g .

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1. Let $$ X $$ be the number of eggs with double yolks in a box of 12. Since each egg is independent with probability $$ p=0.05 $$ of having double yolks,  
   $$
   X \sim \text{Binomial}(n=12,\, p=0.05).
   $$
   The probability that exactly one egg in a box has double yolks is  
   $$
   P(X=1)=\binom{12}{1}(0.05)^1(0.95)^{11}.
   $$

2. A customer buys 3 boxes. Denote by $$ Y $$ the number of boxes (out of 3) that contain exactly 1 egg with double yolk. Since the outcome for each box is independent with probability  
   $$
   p=\binom{12}{1}(0.05)(0.95)^{11}
   $$
   of “success”, we have  
   $$
   Y \sim \text{Binomial}(n=3,\, p).
   $$
   The probability that exactly 2 of the boxes contain exactly 1 egg with double yolk is  
   $$
   P(Y=2)=\binom{3}{2}\Bigl[\binom{12}{1}(0.05)(0.95)^{11}\Bigr]^2\Bigl(1-\binom{12}{1}(0.05)(0.95)^{11}\Bigr).
   $$

3. The farmer delivered 10 boxes (i.e. a total of $$10 \times 12=120$$ eggs) and we want the probability that there are at least 9 eggs with double yolks among these 120 eggs. Let  
   $$
   Z \sim \text{Binomial}(n=120,\, p=0.05)
   $$
   denote the total number of eggs with double yolks. Since $$ np=6 $$ is moderately small, a Poisson approximation is appropriate with  
   $$
   \lambda = np = 6.
   $$
   Under this approximation, we have  
   $$
   P(Z \ge 9) \approx 1-\sum_{k=0}^{8} \frac{6^k e^{-6}}{k!}.
   $$

4. The weight of an individual egg is normally distributed with mean $$ \mu = 65 \,\text{g} $$ and standard deviation $$ \sigma = 2.4 \,\text{g} $$. Let $$ W $$ be the weight of a randomly chosen egg, i.e.  
   $$
   W \sim N(65,\, (2.4)^2).
   $$
   We wish to compute  
   $$
   P(W>68).
   $$
   Standardizing,  
   $$
   Z=\frac{W-65}{2.4},
   $$
   so that  
   $$
   P(W>68)=P\left(Z>\frac{68-65}{2.4}\right)=P\left(Z>1.25\right).
   $$
   From standard normal tables,  
   $$
   P(Z>1.25)\approx 0.1056.
   $$
1. In a box of 12 eggs, the probability of exactly one egg having double yolks is calculated using the binomial probability formula. 2. For a customer buying 3 boxes, the probability that exactly 2 boxes contain exactly one egg with double yolks is determined using the binomial distribution. 3. When the farmer delivers 10 boxes (120 eggs), the probability of having at least 9 eggs with double yolks is approximated using the Poisson distribution. 4. The probability that a randomly chosen egg weighs more than 68 grams is found by standardizing the normal distribution and using standard normal tables.

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