Approximately 3% of the eggs in a store are cracked. If you buy two dozen eggs, what is the probability that no more than two of your eggs are cracked?

Question

Round to 3 decimal places

Answer 1

This is a binomial probabilty problem

 

Let X be the event that your eggs are cracked. P(x) = 0.03  

P(eggs that are not cracked)=1-0.03=0.97

 

P(no more than two of your eggs are cracked)= P(zero eggs are cracked)+ P(1 egg is cracked) +P( 2 eggs are cracked)

                                                                  = P(X=0)+P(X=1)+P(X=2)

 

Now, P(X=0) = C (24,0) (0.03)^0  (0.97)^24  = 0.481

P(X=1) = C (24,1) (0.03)^1 (0.97)^23 = 0.357

P(X=2) = C (24,2) (0.03)^2 (0.97)^22 = 0.127

 

Hence the total is 0.965