Engineering question RE: probability (conditional probability)?

Question

The concept is purely mathematical. Don't worry about the engineering terms! i thought this was easy but i've looked at my answer again and i get 2 different ones. Any help much appriciated. The question is:

A and B represent 2 footings resting on soil. Each footing may remain level OR settle 5cm.

The probability of settlement in each footing is 0.1

HOWEVER, the probability that a footing will settle given that the other has settled is 0.8

WHAT IS THE PROBABILITY OF SETTLEMENT?

Answer 1

0.12
P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)P(B|A) = 0.1 + 0.1 - 0.1(0.8)

Answer 2

i think my argument will not be nicely received. I say the threat is 50% enable a ??, enable b ?? and randomly opt for the values for a and b. As already reported, for a ? 0, P( a < b²) = a million, that is trivial. really somewhat a lot less trivial is the concept that P(a < 0 ) = a million/2 and for this reason P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what occurs even as a > 0 For a > 0, even as that's common to educate there's a non 0 threat for a finite b, the reduce, the threat is 0. a < b² is an identical as putting forward 0 < a < b², remember we are really searching at a > 0. If this a finite period on a limiteless line. The threat that a is an area of this period is 0. P( a < b² | a > 0) = 0 As such we've a complete threat P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 remember, it is because of endless instruments. no count number what style of period you draw on paper or on a pc you'll hit upon a finite threat that looks to mind-set a million. yet it is because of the finite random style turbines on the pc and if we had this question requested with finite values there will be a a answer better than 50%. i do not propose to be condescending, yet please clarify why utilizing the Gaussian to approximate a uniform distribution is a competent theory? are not endless numbers relaxing. Cantor even as mad operating with them! :)

Answer 3

Well the probability of the first one not settling is 0.9. The probability of the second one not settling is also 0.9. Thus the probability of neither settling 0.9x0.9=0.81.
The probability of one settling, then the other settling is 0.1x0.8=0.08. That occurs twice, so it's 0.16
1-0.16-0.81=0.03, which is the probability of one settling, but not the other.

Answer 4

Ok i cant help you but try askin the teacher in your class that might help my friend thinks so to