Question

Suppose that only two factories make Playstation machines. Factory 1 produces 70% of the machines and Factory 2 produces the remaining 30%. Of the machines produced in Factory 1, 2% are defective. Of the machines produced in Factory 2, 5% are defective. What proportion of Playstation machines produced by these two factories are defective? Suppose that you purchase a playstation machine and it is defective. What is the probability that it was produced by Factory 1?

Answer 1

Given:

Factory 1 produces 70%

Factor 2 produces 30%

Defective machines in factory 1 = 2%

Defective machines in factory 2 = 5%

Find-:

What is the probability that it was produced by Factory 1?

Explanation-:

Probability of machines produced by factory1

`\begin{gathered} P(F_1)=70\% \\ \\ P(F_1)=(70)/(100) \\ \\ P(F_1)=(7)/(10) \\ \end{gathered}`

Probability of machines produced by factory 2

`\begin{gathered} P(F_2)=30\% \\ \\ P(F_2)=(30)/(100) \\ \\ P(F_2)=(3)/(10) \end{gathered}`

Probability of factory 1 produced defective item,

`\begin{gathered} P((x)/(F_1))=2\% \\ \\ P((x)/(F_1))=(2)/(100) \\ \\ P((x)/(F_1))=(1)/(50) \end{gathered}`

Probability of factory 2 produced defective item,

`\begin{gathered} P((x)/(F_2))=5\% \\ \\ P((x)/(F_2))=(5)/(100) \\ \\ P((x)/(F_2))=(1)/(20) \end{gathered}`

So, the probability that randomly selected items was form factor 1.

`P((F_1)/(x))\text{ is}`

Now, apply Bayes theorem is:

`P((F_1)/(x))=(P(F_1)P((x)/(F_1)))/(P(F_1)P((x)/(F_1))+P(F_2)P((x)/(F_2)))`

So, the value is:

`\begin{gathered} =((7)/(10)*(1)/(50))/((7)/(10)*(1)/(50)+(3)/(10)*(1)/(20)) \\ \\ =((7)/(500))/((7)/(500)+(3)/(200)) \\ \\ =((7)/(5))/((7)/(5)+(3)/(2)) \\ \\ =((7)/(5))/((14)/(10)+(15)/(10)) \\ \\ =((7)/(5))/((14+15)/(10)) \\ \\ =(7)/(5)*(10)/(29) \\ \\ =(14)/(29) \end{gathered}`

So, the probability is 14/29.