Probability 0.3 0.05 0.5 0.05 (a) P({a, c, e}) P({a, c, e}) = (b) P(E ∪ F), where E = {a, c, e} and F = {b, c, e} P(E ∪ F) = (c) P(E ′), where E is as in part (b) P(E ′) = (d) P(E ∩ F ), where E and F are as in part (b) P(E ∩ F) =
Probability 0.3 0.05 0.5 0.05 (a) P({a, c, e}) P({a, c, e}) = (b) P(E ∪ F), where E = {a, c, e} and F = {b, c, e} P(E ∪ F) = (c) P(E ′), where E is as in part (b) P(E ′) = (d) P(E ∩ F ), where E and F are as in part (b) P(E ∩ F) =
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 11ECP: A manufacturer has determined that a machine averages one faulty unit for every 500 it produces....
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Question
Complete the following probability distribution table and then calculate the stated probabilities.
Outcome | a | b | c | d | e |
---|---|---|---|---|---|
Probability | 0.3 | 0.05 | 0.5 | 0.05 |
(a)
P({a, c, e})
P({a, c, e}) =
(b)
P(E ∪ F), where E = {a, c, e} and F = {b, c, e}
P(E ∪ F) =
(c)
P(E ′), where E is as in part (b)
P(E ′) =
(d)
P(E ∩ F ), where E and F are as in part (b)
P(E ∩ F) =
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