If 6 of 18 new buildings in a city Violate The Building Code, What Is The Probability That a building inspector, Who Randomly selects four of the New buildings for inspection, Will catch
(A) None Of The Buildings That Violate The Building Code;
(B) 1 of the New Buildings That Violate The Building Code;
(C) 2 of the New Buildings That Violate The Building Code;
(D) at least 3 of the New Buildings That Violate The Building Code?
PLEASEHELP ME!
Shirely Amanza says...
This is a BINOMIAL DISTRIBUTION problem.
The probability of exactly x successes is n trials is:
P(X=x)=b(x;n,p) = (nCx)(p^x)((1-p)^(n-x))
where nCx is number of combinations of n things taken x at a time, and "^" means exponentiation.
n = 4
p = 6/18 = .3333
For x = 0,P(X=0) = (4C0)(0.3333^0)(0.6667^4) = 0.1976
xp(X=x)
00.197570373
10.395081479
20.29627
30.098740742
40.012340741
At least 3 means 3 or 4:0.098740742 + 0.012340741 = .1108
Posted on March 29, 2011
