Answer with Step-by-step explanation:
The probability that the device is good drawn in first try
[tex]P(E)=\frac{4}{60}=\frac{1}{15}[/tex]
Probability that the device drawn is defective in first try is
[tex]P(E_2)=\frac{56}{60}=\frac{14}{15}[/tex]
Part a)
Probability that the device obtained in first try is defective is 1/15.
Part b)
Probability that the device obtained in first try is good and the second is defective is
[tex]P(E')=P(E_2)\times P(E_3)[/tex]
where
[tex]P(E_3)[/tex] is the probability that the second is defective provided first is good is [tex]\frac{4}{59}[/tex]
Thus
[tex]P(E')=\frac{56}{60}\times \frac{4}{59}=0.0632[/tex]
Part c)
Probability that all the three samples are good are
Probability that first device is good times
Probability that second device is also good times
Probability that third device is good
[tex]P(E)=\frac{56}{60}\times \frac{55}{59}\times \frac{54}{58}=0.81[/tex]
