Answer:
a. 0.12109
b. 0.0001668
c .0.9726
d. 0.01038
e. 0.01211
f. 0.000001731
Step-by-step explanation:
Sample size = 580
Defective units = 8
Number of picks = 2
a) If the first container is defective, there 7 defective containers left in a population of 579. The probability of selecting a defective one is:
[tex]P=\frac{7}{579}=0.121[/tex]
b) The probability that both are defective is given by:
[tex]P=\frac{8}{580}*\frac{7}{579}= 0.000167[/tex]
c) The probability that both are acceptable is given by:
[tex]P=\frac{580-8}{580}*\frac{579-8}{579}= 0.9726[/tex]
d) In this case, two defective units were removed from the batch, the probability that the third is also defective is:
[tex]P=\frac{6}{578}}= 0.0104[/tex]
e) In this case, one acceptable and one defective unit were removed from the batch, the probability that the third is also defective is:
[tex]P=\frac{7}{578}= 0.01211[/tex]
f) The probability that all three are defective is given by:
[tex]P=\frac{8}{580}*\frac{7}{579}*\frac{6}{578} = 0.000001731[/tex]
