Question 1087613
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