What are the two solutions of the equation? x = StartRoot 5 EndRoot and x = negative StartRoot 5 EndRoot x = 3 + StartRoot 5 End
2 answers:
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Answer:
a) P(E|F) = 0.5
b) P(F|E) = 0.167
c) P(E|F') = 0.625
d) P(E′|F′) = 0.375
Step-by-step explanation:
P(E) = 0.6
P(F) = 0.2
P(E n F) = 0.1
a) P(E|F) = Probability of E occurring, given F has already occurred. It is given mathematically as
P(E|F) = [P(E n F)]/P(F) = 0.1/0.2 = 0.5
b) P(F|E) = Probability of F occurring, given E has already occurred. It is given mathematically as
P(F|E) = [P(E n F)]/P(E) = 0.1/0.6 = 0.167
c) P(E|F′) = Probability of E occurring, given F did not occur. It is given mathematically as
P(E|F') = [P(E n F')]/P(F')
But P(F') = 1 - P(F) = 1 - 0.2 = 0.8
P(E n F') = P(E) - P(E n F) = 0.6 - 0.1 = 0.5
P(E|F') = 0.5/0.8 = 0.625
d) P(E′|F′) = [P(E' n F')]/P(F')
P(F') = 0.8, P(Universal set) = P(U) = 1
P(E' n F') = P(U) - [P(E n F') + P(E' n F) + P(E n F) = 1 - (0.5 + 0.1 + 0.1) = 0.3
P(E′|F′) = 0.3/0.8 = 0.375
