Math 3680 Homework 2 x NGA woman sued a comp x c webassign.net/webStudent/Assign
ID: 3027909 • Letter: M
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Math 3680 Homework 2 x NGA woman sued a comp x c webassign.net/webStudent/Assignment Responses/submit?dep 15568135 The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is o 4s the analogous probability for the second signal is 0.ss, and the probability that he must stop at at least of the two signals B5. a) what is the probability that he must stop at both signals? b) What is the probability that he must stop at the frst sgnal but not at the second one? what is the probability that he must stop at exacty one signal? A family consisting of three persons-A, e, and c-goes to a medical clinic that always has a doctor at each of stations 1, 2, and a, During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of station number for each member suppose that any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other recording the probability that individuals have been assigned. What is the a) All three family members are assigned to the same station? (Round your answer to three decimal places.) 11 gned to the same station? (Round your answer to three decimal places.)Explanation / Answer
Let A be the event of stopping at first signal
Let B be the event of stopping at second signal
Given P(A) =0.45
P(B)=0.55
P(A or B) =0.85
a)probability of stopping at both signals is
P(A and B) =P(A) +P(B) - P(A or B)
=0.45+0.55-0.85
=0.15
b) P(A and Bc) =P(A) - P(A and B)
=0.45-0.15=0.3
c)P(A and Bc) +P(Ac and B) =P(A) P(Bc) +P(Ac) P(B)
=0.45(1-0.55)+(1-0.45)0.55
=0.505