Need help with 4.26 a-d If 6 of 18 new buildings in a city violate the building

Question

Need help with 4.26 a-d

If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch none of the buildings that violate the building code? 1 of the new buildings that violate the building code? 2 of the new buildings that violate the building code? at least 3 of the new buildings that violate the building code? Among the 16 cities that a professional society is considering for its next 3 annual conventions, 7 are in the western part of the United States. To avoid arguments, the selection is left to chance. If none of the cities can be chosen more than once, what are the probabilities that none of the conventions will be held in the western part of the United States? all of the conventions will be held in the western part of the United States?

Explanation / Answer

this is bionomial distribution with

p =6/18
p = 1/3

q = 1- p = 1-1/3 = 2/3
n =4

a)P(x=0) probablity that none of building violate code

= 4 c 0 (1/3)^0 (2/3)^4
= 0.19753086419792592592592622222222

b) P(x=1) probablity that one of building violate code

= 4 C 1 (1/3)^1 (2/3)^3
= 0.39506172839525925925925896296295

c) P(x=1) probablity that two of building violate code

= 4 C 2 (1/3)^2 (2/3)^2
= 0.29629629629599999999999977777774

d) P(x>=3)
P(x=3) + P(x=4)
= 4 c 3 (1/3)^3 (2/3) + 4c4 (1/3)^4
= 4 (1/3)^3(2/3) + (1/3)^4
=0.11111111111081481481481503703701

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