study guide - P(x)= 0<x<1 IN A PROBABILITY PROBLEM IF THEY...

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Neither Just one Just one Both P(x)= 0<x<1 IN A PROBABILITY PROBLEM IF THEY ASK YOU FOR JUST ONE OR EXACTLY ONE OUT OF A CERTAIN NUMBER OF TRIALS, MAKE SURE TO ALTERNATE IT. Could also account for this by multiplying by factorial o Ex. 3 possible defective and 17 regular, choose just one defective o [(3/20) * (17/19) * (16/18)] + [(17/20)*(16/19)*(3/18)]+ [(17/20)*(3/19)*(16/18) IF MUTUALLY EXCLUSIVE IT MEANS THEY ARE INDEPENDENT AND DO NOT AFFECT ONE ANOTHER. THEREFORE, CONDITIONAL PROBABILTY DOES NOT APPLY AND ANSWER IS 0 Can always check probability by doing counting technique Grabbing 2 marbles out of bag without replacement, both numerator and denominator decrease OR RULE: Addition, Either P(A or B)= P(A) + P(B) – P(A and B)….If not mutually exclusive (overlap) P(A or B) either + P(A’ or B’) neither= 1 P(A or B)= P(A and B) + P(A’ and B) + P(A and B’) P(EXACTLY ONE)= P( A or B) – P(A and B) 1-none=either at least one = either= 1-none(opposite) Just one= add up individual probabilities (excluding both) Exactly x- just the individual probability minus the both Conditional Probability: P(A|B)= P(A and B) / P(B) Only use if it’s a given problem that is not multistep Multiplication Rule: And, Both A B A and B’ A’ and B
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P(A and B)= P(A) * P(B) if independent P(A and B)= P(A|B)*P(B) or also P(B|A)* P(A) Independent Events: one event has no impact on the other P(A) * P(B) =? P(A and B) Bayes Rule:
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