Over the past 200 working days, the number of defective parts produced by a machine is given in the following table :
`|{:("Number of","0 1 2 3 4 5 6 7 8 9 10 11 12 13"),("defective parts",),("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.

Explanation

Total number of past working days are 200.
(i) Let `E_(1)` be the event that the tomorrow's output will have no defective part.
We see that the number of times the event happened = 50
`therefore P_(E_(1))=("Number of trials in which the event happened")/("Total number of trials")=(50)/(200)=(1)/(4)=0.24`
(ii) Let `E_(2)` be the event that the tomorrow's output will have at least one defective part.
We see that the number of times the event happened
= `200-"(Number of no defective parts)"`
`=200-20=150`
`thereforeP(E_(2))=("Number of trials in which the event happpened")/("Total number of trials")=(150)/(200)=(3)/(4)=0.75`
(iii) Let `E_(3)` be the event that the tomorrow's output will have not more than 5 defective parts.
We see that the number of times the event happened, i.e., no defective part, one defective part, two defective parts, three defective parts, four defective parts and five defective parts
`=50+32+22+18+12+12=146`
`thereforeP(E_(3))=("Number of trials in which the event happened")/("Total number of trials")=(146)/(200)=0.73`
(iv) Let `E_(4)` be the event that tomorrow's output will have more than 13 defective parts.
We see that the number of times the event happened = 0
(since there are no more than 13 defective parts)
`thereforeP(E_(4))=("Number of trials in which the event happened")/("Total number of trials")=(0)/(200)=0`