Let events A,B and C are
A : Bolt is manufactured by machine A.
B : Bolt is manufactured by machine B.
C : Bolt is manufactured by machine C.
P(A) = 25% = \(\frac{25}{100}\) = \(\frac{1}{4}\)
P(B) = 35% = \(\frac{35}{100} = \frac{7}{20}\)
P(C) = 40% = \(\frac{40}{100} = \frac{2}{5}\)
let E be event that drawn bold in defective.
P(E/A) = 5% = \(\frac{5}{100} = \frac{1}{20}\)
P(E/B) = 4% = \(\frac{4}{100} = \frac{1}{25}\)
P(E/C) = 2% = \(\frac{2}{100} = \frac{1}{20}.\)
Probability that defective bolt is manufactured by C
is P(C/E) = \(\frac{P(E/C)P(C)}{P(E/C).P(C)+P(E/B)\, P(B)+P(E/A)P(A)}\)
= \(\cfrac{\frac{1}{50} \times \frac{2}{5}}{\frac{1}{50}\times \frac{2}{5} + \frac{1}{25} \times \frac{7}{20}+ \frac{1}{20}\times\frac{1}{4}}\)
=\(\cfrac{\frac{1}{125}}{\frac{1}{125} + \frac{7}{500}+\frac{1}{80}}\)
= \(\cfrac{\frac{1}{125}}{\frac{16+28+25}{2000}}\)
= \(\frac{16}{69}\)
Thus probability that defective bolt is not manufactured by C
is P(C'/E) = 1-P(C/E)
= 1 - \(\frac{16}{69}\) = \(\frac{69-16}{69}\) = \(\frac{53}{69}\).