SOLUTION: A really bad carton of 18 eggs contains 7 spoiled eggs. An unsuspecting chef picks 4 eggs at random for his �Mega-Omelet Surprise.� Find the probability that the number of unspoil

Question 758179: A really bad carton of 18 eggs contains 7 spoiled eggs. An unsuspecting chef picks 4 eggs at random
for his �Mega-Omelet Surprise.� Find the probability that the number of unspoiled eggs among the
4 selected is
a. exactly 4 b. 2 or fewer
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
n = 18
7 are spoiled
chef picks 4 out of the 18.
possible combinations are 18C4 = 3060

probability that he will get 4 spoiled eggs is equal to 7C4 / 18C4 which is equal to 35 / 3060.

probabiltiy that he can get 0 spoiled eggs is equal to 11C4 / 3060 which is equal to 330 / 3060.

probabiltiy that he can get 1 spoiled egg is equal to (7C1 * 11C3) / 18C4 which is equal to (7 * 165) / 3060 which is equal to 1155 / 3060.

probabiltiy that he can get 2 spoiled eggs is equal to (7C2 * (11C2) / 18C4 which is equal to (21 * 55) / 3060 which is equal to 1155 / 3060.

probabiltiy that he can get 0 or 1 or 2 spoiled eggs is equal to (330 + 1155 + 1155) / 3060 which is equal to 2640 / 3060.

this is impossible to visualize with so many eggs.
reduce the problem and we may be able to visualize how the solution is obtained.
suppose 5 eggs of which 3 are spoiled.

chef draws 2 eggs.
what is the probability that both are bad?
what is the probability that 0 or 1 are bad?

probability of both eggs being spoiled would be equal to:

3C2 / 5C2 which is equal to 3 / 10

label the eggs:

s1
s2
s3
g1
g2

possible combinations of 2 eggs out of 5 is shown below:

s1, s2 ***** 2 eggs spoiled
s1, s3 ***** 2 eggs spoiled
s1, g1
s1, g2
s2, s3 ***** 2 eggs spoiled
s2, g1
s2, g2
s3, g1
s3, g2
g1, g2

the formula works for 2 spoiled eggs.

probability of getting 0 spoiled eggs is equal to 2C2 / 5C2 which is equal to 1 / 10.

the total possible combinations are, once again:

s1, s2
s1, s3
s1, g1
s1, g2
s2, s3
s2, g1
s2, g2
s3, g1
s3, g2
g1, g2 ***** 0 eggs spoiled

probability of getting 0 spoiled eggs is 1 out of 10.

probability of getting 1 spoiled egg is equal to (3C1 * 2C1) / 5C2 which is equal to (3 * 2) / 10 which is equal to 6 / 10

the total possible combinations are, once again:

s1, s2
s1, s3
s1, g1 ***** 1 egg spoiled
s1, g2 ***** 1 egg spoiled
s2, s3
s2, g1 ***** 1 egg spoiled
s2, g2 ***** 1 egg spoiled
s3, g1 ***** 1 egg spoiled
s3, g2 ***** 1 egg spoiled
g1, g2

the formulas can be demonstrated to work with small numbers.
this isn't possible with big numbers.
some logic needs to be applied to convince yourself that the same formulas will work with bigger number.
the rest is on faith that your logic is good.
probability of getting 0 or 1 spoiled egg out of 5 is equal to 1/10 + 6/10 which is equal to 7/10 as shown below:

s1, s2
s1, s3
s1, g1 ***** 1 egg spoiled
s1, g2 ***** 1 egg spoiled
s2, s3
s2, g1 ***** 1 egg spoiled
s2, g2 ***** 1 egg spoiled
s3, g1 ***** 1 egg spoiled
s3, g2 ***** 1 egg spoiled
g1, g2 ***** 0 eggs spoiled

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