What are the domain and range of the function below?
2 answers:
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Answer:
1) Probability that all five are good = 0.46
2) P(at most 2 defective) = 0.99
3) Pr(at least 1 defective) = 0.54
Step-by-step explanation:
The total number of bulbs = 30
Number of defective bulbs = 4
Number of good bulbs = 30 - 4 = 26
Number of ways of selecting 5 bulbs from 30 bulbs =
ways
Number of ways of selecting 5 good bulbs from 26 bulbs =
ways
Probability that all five are good = 65780/142506
Probability that all five are good = 0.46
2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)
Pr(no defective) has been calculated above = 0.46
Pr(1 defective) =
Pr(1 defective) = (14950*4)/142506
Pr(1 defective) =0.42
Pr(2 defective) =
Pr(2 defective) = (2600 *6)/142506
Pr(2 defective) = 0.11
P(at most 2 defective) = 0.46 + 0.42 + 0.11
P(at most 2 defective) = 0.99
3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)
Pr(1 defective) =0.42
Pr(2 defective) = 0.11
Pr(3 defective) =
Pr(3 defective) = 0.009
Pr(4 defective) =
Pr(4 defective) = 0.00018
Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018
Pr(at least 1 defective) = 0.54