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

Determine the probability that tomorrow's output will have

Options

A. no defective part.
B. atleast one defective part.
C. nor more than 5 defective parts.
D. more than 13 defective parts.

Explanation

Total numbers of working days, n(S)=200
i) Number of days in which no defective parts = `(n(E_(1)))/(n(S))= 50/200 = 1/4 = 0.25`
Hence, the probability that no defective part is 0.25.
ii) Number of days in which atleast one defective part, is
`n(E_(2)= 32+22+18+12+12+10+10+8+6+6+2+2=150`
`therefore` Probability that atleast one defective part `=(n(E_(2)))/(n(S))= 150/200=3/4=0.75`
Hence, the probability that all least one defective part is 0.75.
iii) Number of days in which not more than 5 defective parts,
`n(E_(3))=50+32+22+12=146`
`therefore` Probability that not more than 5 defective parts
`(n(E_(3)))/(n(S))=146/200=0.73`
Hence, the probability that not more than 5 defective parts is 0.73.
iv) Number of days in which more than 13 defective parts
`n(E_(4))=0`
`therefore` Probability that more than 13 defective parts
`=(n(E_(4)))/(n(S))==0/200=0`
Hence, the probability that more than 13 defective parts is 0.