https://cis.temple.edu/~tug29203/18spring-3329/reading/hw4a.pdf
5. (2 points) Consider the SDN OpenFlow network shown in the following gure. Suppose that the desired forwarding behavior for datagrams arriving from host h3 or h4 at s2 is as follows: any datagrams arriving from host h3 and destined for h1, h2, h5 or h6 should be forwarded OpenFlow)example) should be forwarded Host h6
https://www.fox.temple.edu/sites/fox/files/smj2112-Luo.pdf
growth V13.Firm 0.136 − 0.0970.0050.0840.0880.0270.0180.0070.0160.0260.0630.1271.000 advertising V14.Manufacturing0.186 − 0.095 − 0.016 − 0.031 − 0.026 − 0.008 − 0.024 − 0.0180.0120.0230.052 − 0.0780.0271.000 V15.Numberof 0.005 − 0.038 − 0.010 − 0.051 − 0.033 − 0.0270.0030.0050.0030.0070.008 − 0.0160.0240.0311.000
https://cis.temple.edu/~latecki/Courses/CIS2033-Spring12/ElementaryProbabilityforApplications/ch3.pdf
The first step in analyzing craps is to compute the probability that the player makes his point. Suppose his point is 5 and let Ek be the event that the sum is k. There are 4 outcomes in E5 ((1, 4), (2, 3), (3, 2), (4, 1)), 6 in E7, and hence 26 not in E5 ∪ E7. Letting × stand for “the sum is not 5 or 7,” we see that 4 26 4 = P(5) P(× 5