If drawn bolt is defective , then the probability that it is not manufactured by C is ?

Question

In a bolt factory, machines A, B, and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5%, 4% and 2% are respectively defective bolt. On the basis of the above information answer the following, 

 If drawn bolt is defective , then the probability that it is not manufactured by C is
(a) \(\frac{39}{69}\)   (b) \(\frac{40}{69}\)   (c) \(\frac{41}{69}\)   (d) \(\frac{42}{69}\)

My answer is coming to be \(\frac{53}{69} ​\), which is not in the options. Please give CBSE Boards oriented solution with steps. 


 

Answer 1

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 - \(\frac{16}{69}\) = \(\frac{69-16}{69}\) = \(\frac{53}{69}\).