In a bolt factory, machines A, B and C manufacture respectively 20...

Let $X ≡$Event that product is defective, then

$P\left(\dfrac{X}{A}\right) = \dfrac{3}{100}P\left(\dfrac{X}{B}\right) = \dfrac{4}{100}P\left(\dfrac{X}{C}\right) = \dfrac{2}{100}$

Now,

$P\left(\dfrac{C}{X}\right) = \dfrac{P\left(C\right)P\left(\dfrac{X}{C}\right)}{P\left(A\right)P\left(\dfrac{X}{A}\right) + P\left(B\right)P\left(\dfrac{X}{B}\right) + P\left(C\right)P\left(\dfrac{X}{C}\right)}$

$⇒ P\left(\dfrac{C}{X}\right) = \dfrac{\dfrac{50}{100} \times \dfrac{2}{100}}{\dfrac{20}{100} \times \dfrac{3}{100} + \dfrac{30}{100} \times \dfrac{4}{100} + \dfrac{50}{100} \times \dfrac{2}{100}}$

$⇒ P\left(\dfrac{C}{X}\right) = \dfrac{100}{60 + 120 + 100}$

$⇒ P\left(\dfrac{C}{X}\right) = \dfrac{5}{14}$

Hence this is the correct option.