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  • Over the past 200 working days, the number of defective parts produced by a machine is given in the following table: Determine the probability that tomorrow’s output will have (i) no defective part (ii) atleast one defective part (iii) not more than 5

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:

Number of defective parts 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Days 50 32 22 18 12 12 10 10 10 8 6 6 2 2

Determine the probability that tomorrow’s output will have

(i) no defective part

(ii) at least one defective part

(iii) not more than 5 defective parts

(iv) more than 13 defective parts

Answers (3)

\text{Probability is defined as}=\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events = total number of working days =200

(i) no defective part

Favourable outcomes =50 days =50

\text{p (no defective part)}=\frac{50}{200}=0.25

(ii) Probability that at least one defective part = 1 - the probability that no defective part

\text{p (no defective part)}=\frac{50}{200}=0.25

\text{p (at least one defective part)}=1-\frac{50}{200}=0.75

(iii) not more than 5 defective parts = 0 or 1 or 2 or 3 or 4 or 5 defective parts

p(not more than 5 defective parts) = p(no. defective part) + p(1 defective part) + p(2 defective part) + p(3 defective part) + p(4 defective part) + p(5 defective part)

=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}

=\frac{50+32+22+18+12+12}{200}=\frac{146}{200}  

=\frac{73}{200}=0.73                     

(iv) more than 13 defective parts = not possible

Favourable outcomes = 0

p (more than 13 defective parts) = 0

Posted by

infoexpert23

View full answer

Probability is defined as =\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events = total number of working days = 200

(i) no defective part

Favourable outcomes = 50 days = 50

p (no defective part)=\frac{50}{200}=0.25

(ii) Probability that at least one defective part = 1 - probability that no defective part

p (no defective part)=\frac{50}{200}=0.25

p (at least one defective part) =1-\frac{50}{200}=0.75

(iii) not more than 5 defective parts = 0 or 1 or 2 or 3 or 4 or 5 defective parts

p(not more than 5 defective parts) = p(no. defective part) + p(1 defective part) + p(2 defective part) + p(3 defective part) + p(4 defective part) + p(5 defective part)

=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}

=\frac{50+32+22+18+12+12}{200}=\frac{146}{200}  

=\frac{73}{200}=0.73                     

(iv) more than 13 defective parts = not possible

Favourable outcomes = 0

p (more than 13 defective parts) = 0

Posted by

infoexpert23

View full answer

Probability is defined as =\frac{\text {Favourable outcomes}}{\text {Total number of events}}

Here, total events = total number of working days = 200

(i) no defective part

Favourable outcomes = 50 days = 50

p (no defective part)=\frac{50}{200}=0.25

(ii) Probability that at least one defective part = 1 - probability that no defective part

p (no defective part)=\frac{50}{200}=0.25

p (at least one defective part) =1-\frac{50}{200}=0.75

(iii) not more than 5 defective parts = 0 or 1 or 2 or 3 or 4 or 5 defective parts

p(not more than 5 defective parts) = p(no. defective part) + p(1 defective part) + p(2 defective part) + p(3 defective part) + p(4 defective part) + p(5 defective part)

=\frac{50}{200}+\frac{32}{200}+\frac{22}{200}+\frac{18}{200}+\frac{12}{200}+\frac{12}{200}

=\frac{50+32+22+18+12+12}{200}=\frac{146}{200}  

=\frac{73}{200}=0.73                     

(iv) more than 13 defective parts = not possible

Favourable outcomes = 0

p (more than 13 defective parts) = 0

Posted by

infoexpert23

View full answer

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