Question

A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p is

(1) 54 

(2) 64 

(3) 66 

(4) 56

Answer 1

Correct option is (1) 54  

	A	B
Manufactured	60%	40%
Standard quality	80%	90%

P(Manufactured at B / found standard quality) = ? 

A : Found S.Q 

B : Manufacture B 

C : Manufacture A

\( P\left(E_{1}\right)=\frac{40}{100}\)

\( \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{60}{100}\)

\(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=\frac{90}{100}\)

\(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=\frac{80}{100}\)

\(\because \mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{E}_{2}\right)}=\frac{3}{7}\)

\(\therefore 126 \mathrm{P}=54\)