A computer producing factory has only two plants T _ 1 and T _ 2 ....

$P\left({T}_{1}\right) = \dfrac{20}{100}$

$P\left({T}_{2}\right) = \dfrac{80}{100}$

Let, $P\left(\dfrac{D}{{T}_{2}}\right) = x$ (where, $D$ represents defective units)

$P\left(\dfrac{D}{{T}_{1}}\right) = 10x$

$P\left(D\right) = \dfrac{7}{100}$ (given)

$P\left({T}_{1}\right)P\left(\dfrac{D}{{T}_{1}}\right) + P\left({T}_{2}\right)P\left(\dfrac{D}{{T}_{2}}\right) = \dfrac{7}{100}$

$\dfrac{20}{100} \times 10x + \dfrac{80}{100} \times x = \dfrac{7}{100}$

$x = \dfrac{1}{40}$

$P\left(\dfrac{D}{{T}_{2}}\right) = \dfrac{1}{40}$

$⇒ P\left(\overset{-}{D}{T}_{2}\right) = \dfrac{39}{40} =$Probability of not defective, given that it is produced in plant ${T}_{2}$

$P\left(\dfrac{D}{{T}_{1}}\right) = \dfrac{10}{40} ⇒ P\left(\overset{-}{D}{T}_{1}\right) = \dfrac{30}{40}$

Now using Bayes' theorem

Probability of computer from ${T}_{2}$ given that it is not defective :

$P\left({T}_{2}\overset{-}{D}\right) = \dfrac{\dfrac{80}{100} \times \dfrac{39}{40}}{\dfrac{20}{100} \times \dfrac{30}{40} + \dfrac{80}{100} \times \dfrac{39}{40}} = \dfrac{78}{93}$