Answer:
(a) 0.0833 or 8.33%
(b) 0.40 or 40%
Step-by-step explanation:
Parts line one (n1) = 1,000 parts
Defects line one (d1) = 100 parts
Parts line two (n2) = 2,000 parts
Defects line two (d2) = 150 parts
Total number of parts (n) = 3,000 parts
a. Probability of a randomly selected part being defective:
[tex]P(d) = \frac{d_1+d_2}{n_1+n_2}=\frac{100+150}{1,000+2,000}\\P(d) =0.0833=8.33\%[/tex]
The probability is 0.0833 or 8.33%
b. Probability of a part being produced by line one, given that it is defective:
[tex]P(1|d)=\frac{P(1\cap d)}{P(1\cap d)+P(2\cap d)}\\P(1|d)=\frac{\frac{100}{3,000} }{\frac{100}{3,000}+\frac{150}{3,000}}\\P(1|d)=0.4 = 40\%[/tex]
The probability is 0.40 or 40%.