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Mathematics 13 Online
OpenStudy (anonymous):

three highways connect city A with city B.two highways connect city B with city C.during rush hours,each highway is blocked by a traffic accident with probability 0.2,independently of other highways. (a) compute the probability that there is atleast one open route from A to C.

OpenStudy (anonymous):

P(me knowing how to do this) = 0

OpenStudy (hoblos):

P(at least one open route from A to C) = 1-P( no open route from A to C) P( no open route from A to C) = P( (no open route from A to B)U(no open route from B to C) = (0.2)^3 + (0.2)^2 = 0.048 P(at least one open route from A to C) = 1-0.048=0.952

OpenStudy (anonymous):

thank you.well explained :)

OpenStudy (hoblos):

my pleasure :)

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