Question 1087613: A basket contains 9 eggs, 3 of which are cracked. If we randomly select 4 of the eggs for hard boiling, what is the probability of the following events?
a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question:
A basket contains 9 eggs, 3 of which are cracked. If we randomly select 4 of the eggs for hard boiling, what is the probability of the following events?
a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.
Solution:
This situation can be solved using the hypergeometric distribution, namely
sampling without replacement for a small population.
Let
C=number of cracked eggs in basket=3
c=number of cracked eggs in sample
U=number of uncracked eggs in basket=6
u=number of uncracked eggs in sample
=> c+u=sample size=4, and C+U=population size=3+6=9
then
P(c,u)=C(C,c)*C(U,u)/C(C+U,c+u)
where
C(N,x) is number of combinations of selecting x objects out of N.
(a) all cracked eggs are selected
P(3,1)=C(3,3)*C(6,1)/C(9,4)=0.0476
(b) none of the cracked eggs are selected
P(0,4)=C(3,0)*C(6,4)/C(9,4)=0.1190
(c) two cracked eggs out of 4
P(2,2)=C(3,2)*C(6,2)/C(9,4)=0.3571
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