Conditional Probability

[Win_odd Dhamnekar]

My attempt to answer (a):
Let us assume P(Guilty) =1 and P(Inguilty) =0.

So,


Author S. M. Ross ( Professor (Industrial engineering and Operations Research), University of Southern California) ) gave the answer [imath]\frac{97}{142}[/imath]

So, would any member of this forum tell me what is wrong with my computation of [imath]\frac{351}{530}[/imath] as an answer satisfactorily?

 

[Dr.Peterson]

View attachment 34128
My attempt to answer (a):
Let us assume P(Guilty) =1 and P(Inguilty) =0.

So,
View attachment 34129

Author S. M. Ross ( Professor (Industrial engineering and Operations Research), University of Southern California) ) gave the answer [imath]\frac{97}{142}[/imath]

So, would any member of this forum tell me what is wrong with my computation of [imath]\frac{351}{530}[/imath] as an answer satisfactorily?

Please explain your calculation. Why didn't you use the fact that 70% of defendants are guilty? You can't assume P(Guilty) =1 when you are told that P(Guilty) =0.7!

 

[Win_odd Dhamnekar]

Please explain your calculation. Why didn't you use the fact that 70% of defendants are guilty? You can't assume P(Guilty) =1 when you are told that P(Guilty) =0.7!

My answer to question (b) :[math] \frac{0.7 ((1-0.7) \times (0.7)^2 + (1-0.7) \times (0.7)^2 + (1-0.2) \times (0.2)^2 +(1-0.2) \times (0.2)^2)}{ 0.7 ((1-0.7) \times (0.7)^2 + (1-0.7) \times (0.7)^2 + (1-0.2) \times (0.2)^2 +(1-0.2) \times (0.2)^2) + 0.3( (0.7) \times (1-0.7)^2 + (0.7) \times (1-0.7)^2 + (0.2) \times (1-0.2)^2 + (0.2) \times (1-0.2)^2 )}= \frac{0.2506}{0.3652}= 0.6862 [/math]
But Author S. M. Ross (Professor, (Industrial engineering and Operations Research), University of Southern California) gave answer [imath] \frac{15}{26}[/imath]

How is that?

Where I am wrong?

 

[Win_odd Dhamnekar]

My answer to question (b) :[math] \frac{0.7 ((1-0.7) \times (0.7)^2 + (1-0.7) \times (0.7)^2 + (1-0.2) \times (0.2)^2 +(1-0.2) \times (0.2)^2)}{ 0.7 ((1-0.7) \times (0.7)^2 + (1-0.7) \times (0.7)^2 + (1-0.2) \times (0.2)^2 +(1-0.2) \times (0.2)^2) + 0.3( (0.7) \times (1-0.7)^2 + (0.7) \times (1-0.7)^2 + (0.2) \times (1-0.2)^2 + (0.2) \times (1-0.2)^2 )}= \frac{0.2506}{0.3652}= 0.6862 [/math]
But Author S. M. Ross (Professor, (Industrial engineering and Operations Research), University of Southern California) gave answer [imath] \frac{15}{26}[/imath]

How is that?

Where I am wrong?

I got the answer as given by author for question (b).