Answer to Question #239489 in Statistics and Probability for James

Question

A box of a dozen eggs contains 7 good eggs and 5 bad eggs. Mr. Thomas Cook is preparing breakfast for his family - one wife and two kids. He plans to cook an egg for each one of them plus some bacon. He randomly selects 4 eggs from the box and sets these aside in a bowl.

a.     Determine the probability distribution function of the number of good eggs contained in the bowl.

b.    What is the probability that the bowl contains at most one bad egg?

c.     What is the probability that Mr. Cook will have to get eggs from the box again?

 

Answer 1

a) 7 good eggs and and 5 bad 

4 eggs are selected

X -number of good eggs 

P(X = k) = (7Ck * 5C(4-k)) /(12C4)  for k = 0,1,..4 

b) probability that there is at most one bad egg 

= probability that there is at least 3 good eggs

=P(X =3) +P(X =4) 

=(7C3*5C1 + 7C4 )/(12C4)

= (175 +35)/495 = 0.42424242

c) probability that Mr Cook will have to get eggs form box again = probability that there is more than one bad egg 

=1- probability that there is at most one bad egg 

= 0.5757575