Question 1197325: Five of 10 new buildings in a city violate the building code. A random sample of three buildings
selected for inspection.
a)What is the probability of none of the buildings violate the building code in a sample of 3 buildings
b)What is the probability of at least one of the new buildings violate the building code in a sample of 3 buildings?
Found 3 solutions by Boreal, math_tutor2020, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Probability the first won't violate is 5/10, then 4/9 then 3/8. That product is 60/720 or 1/12.
At least one of them is 1-1/12 or 11/12. Notice one doesn't have to compute 1 building, 2 buildings or 3 buildings.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Part (a)
5 buildings violate code, 10-5 = 5 do not
5/10 = probability first selection doesn't violate code
4/9 = probability second selection doesn't violate code
3/8 = probability third selection doesn't violate code
The numerators decrease: 5,4,3
So do the denominators: 10,9,8
This assumes that we're sampling without replacement.
Multiply out the fractions to get this result
(5/10)*(4/9)*(3/8) = (5*4*3)/(10*9*8)
(5/10)*(4/9)*(3/8) = (5*4*3)/(2*5*3*3*2*4)
(5/10)*(4/9)*(3/8) = 1/(2*3*2)
(5/10)*(4/9)*(3/8) = 1/12
The probability of selecting 3 buildings that don't violate code is 1/12.
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Alternate route:
n = 5 buildings that don't violate code
r = 3 selections
Use the nCr combination formula
n C r = (n!)/(r!(n-r)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = (20)/(2)
5 C 3 = 10
There are 10 ways to pick three out of the five buildings that are to code.
Using that same formula, you should find that 10C3 = 120 which represents the number of ways to pick any three buildings (whether they are to code or not).
As you can probably see, we are using nCr instead of nPr because order doesn't matter. A grouping like ABC is the same as CBA.
We have 10 ways to pick the three buildings that are to code out of 120 ways to pick the three buildings.
10/120 = 1/12
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Answer: 1/12
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Part (b)
We'll use the answer of the previous part.
The events "all 3 don't violate code" and "at least one violates code" are complementary events.
One or the other must happen.
P(all 3 don't violate code) + P(at least one violates code) = 1
P(at least one violates code) = 1 - P(all 3 don't violate code)
P(at least one violates code) = 1 - 1/12
P(at least one violates code) = 12/12 - 1/12
P(at least one violates code) = (12 - 1)/12
P(at least one violates code) = 11/12
Answer: 11/12
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Five of 10 new buildings in a city violate the building code. A random sample of three  buildings
 selected for inspection.
a) What is the probability of none of the buildings violate the building code in a sample of 3 buildings
b) What is the probability of at least one of the new buildings violate the building code in a sample of 3 buildings?
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The solution of two other tutors is correct, if these 10 buildings are all new buildings in the city,
and there are no other new buildings in the city.
But an opposite assumption can be made, too, that the number of new buildings in the city is very large,
for example, that there are 10,000 new buildings in the city.
Then the condition simply means that the probability that a randomly selected new building in the city violates the code is 1/2 = 0.5.
Then the problem allows and requires different treatment: it can be considered as a standard binomial distribution problem.
From the problem's formulation, it is impossible to determine which one of the two possible treatments is preferable,
so the condition is AMBIGUOUS.
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