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p0941
Posts: 95
Posted 00:55 Aug 25, 2009 |

Ex. 16.

a)      P(HeartDisease=Yes|Diet=Healthy) = P(HeartDeiseas=Yes|Diet=Healthy, Exercise=Yes)+ P(HeartDeiseas=Yes|Diet=Healthy, Exercise=No) = 0.25*0.25 + 0.25*0.55 = 0.2

b)      P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes) = P(HeartDisease=Yes| BloodPressure=High) * P(HeartDisease=Yes| ChestPain=Yes)

= 0.80 (From lecture example 2) * P(ChestPain=Yes|HeartDisease=Y) *P(HeartDisease=Yes) / P(ChestPain=Yes)

= 0.8*0.361375*0.49/0.5175625 = 0.2737 

P.S.

P(HeartBurn=Yes) = 0.2*0.25+0.85*0.75= 0.6875

P(ChestPain=Yes) = 0.8*P(HeartDisease=Yes)*P(HeartBurn=Yes) + 0.6* P(HeartDisease=Yes)*P(HeartBurn=No) + 0.4* P(HeartDisease=Yes)*P(HeartBurn=Yes) + 0.1*P(HeartDisease=Yes)*P(HeartBurn=Yes) = 0.8*(0.49)* 0.6875 + 0.6*(0.49(1-0.6875)) + 0.4*(1-0.49)*0.6875 + 0.1*(1-0.49)* (1-0.6875) = 0.5175625

P(ChestPain=Yes|HeartDisease=Y) = (P(ChestPain=Yes|HeartDiesease=Y , HeartBurn=Yes) + P(ChestPain=Yes|HeartDisease=Y , HeartBurn=No) = 0.8*(0.49)* 0.6875 + 0.6*(0.49(1-0.6875)) = 0.361375

    

p0941
Posts: 95
Posted 12:44 Aug 25, 2009 |

Update:

b)

P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes)
= P(HD=Yes,BP=High,CP=Yes) / P(BP=H, CP=Y)
= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)
= 0.85*0.49 / 0.85*0.49 + 0.2*0.51 = 0.8

P(HD=Yes) = 0.49, from Lecture example -1

 

xieguahu
Posts: 50
Posted 18:30 Aug 27, 2009 |

(a) P(HeartDisease=Yes|Diet=Healthy)

                = p(HeartDisease = Yes|Diet = Healthy, Exercise = Yes)P(Exercise = Yes)

   + P(HeartDisease = Yes | Diet = Healthy, Exercise = No) p(Exercise = No)

= 0.25 X 0.7 +  0.55 X 0.3 = 0.175 + 0.165 = 0.34

 

(b) is the same

 

Last edited by xieguahu at 18:30 Aug 27, 2009.
abhishek_sharma
Posts: 79
Posted 14:55 Aug 29, 2009 |

(a) P(HeartDisease=Yes|Diet=Healthy)

                = p(HeartDisease = Yes|Diet = Healthy, Exercise = Yes)P(Diet=Healthy)P(Exercise = Yes)

   + P(HeartDisease = Yes | Diet = Healthy, Exercise = No)P(Diet=Healthy)P(Exercise = No)

= 0.25 X0.25X 0.7 +  0.55 X0.25 X 0.3 = 0.04375 + 0.04125 = 0.085

 

I think we have to consider P(Diet=Healthy) also.. So I multiplied that value

 

Last edited by abhishek_sharma at 14:56 Aug 29, 2009.
HelloWorld
Posts: 88
Posted 22:06 Aug 29, 2009 |
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

P(HD=Yes | BP= High, CP=Yes)

= P(HD=Yes, BP=High, CP=Yes) / P(BP=High, CP=Yes)

= P(BP=High | HD=Yes, CP=Yes) P(HD=Yes, CP=Yes) / P(BP=High)P(CP=Yes)

and i'm not sure what's the next step..

HelloWorld
Posts: 88
Posted 09:13 Aug 30, 2009 |

Check out my solution

Attachments:
p0941
Posts: 95
Posted 13:38 Aug 30, 2009 |
HelloWorld wrote:
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

 

Because BP and HD are conditionally dependent and they are both conditionally independent on CP.

HelloWorld
Posts: 88
Posted 13:47 Aug 30, 2009 |
p0941 wrote:
HelloWorld wrote:
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

 

Because BP and HD are conditionally dependent and they are both conditionally independent on CP.

CP is dependent on HD though..

xieguahu
Posts: 50
Posted 15:32 Aug 30, 2009 |
I think P(HeartDisease=Yes|Diet=Healthy) means if given diet is healty, what is the percentage of HeartDiseas = Yes,  so we donot need to consider p(Diet = Healthy) in this case.
abhishek_sharma wrote:

(a) P(HeartDisease=Yes|Diet=Healthy)

                = p(HeartDisease = Yes|Diet = Healthy, Exercise = Yes)P(Diet=Healthy)P(Exercise = Yes)

   + P(HeartDisease = Yes | Diet = Healthy, Exercise = No)P(Diet=Healthy)P(Exercise = No)

= 0.25 X0.25X 0.7 +  0.55 X0.25 X 0.3 = 0.04375 + 0.04125 = 0.085

 

I think we have to consider P(Diet=Healthy) also.. So I multiplied that value

 

 

p0941
Posts: 95
Posted 16:13 Aug 30, 2009 |
HelloWorld wrote:
p0941 wrote:
HelloWorld wrote:
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

 

Because BP and HD are conditionally dependent and they are both conditionally independent on CP.

CP is dependent on HD though..

Good point. I update part b here:

 

P(Hb=Y) = P(Hb|D=H)P(D=H) +  P(Hb|D=UnHealthy)(1-P(D=H))= 0.6875
P(Hb=N) = 1-0.6875 = 0.3125

P(CP=Y) = P(CP=Y|HD=Y,Hb=Y)P(HD=Y,Hb=Y)+
          P(CP=Y|HD=Y,Hb=N)P(HD=Y,Hb=N)+
          P(CP=Y|HD=N,Hb=Y)P(HD=N,Hb=Y)+
          P(CP=Y|HD=N,Hb=N)P(HD=N,Hb=N) = 0.51
P(CP=Y|HD=Y) = 0.8+0.6=1.4
P(HD=Yes) = 0.49, from Lecture example -1

P(HD=Y|CP=H) = P(CP=Y|HD=Y)P(HD=Y)/P(CP=Y) = 1.4*0.49 / 0.51 = 1.34


P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes)
= P(HD=Yes,BP=High,CP=Yes) / P(BP=H, CP=Y)
= P(BP=High|HD=Yes)P(HD=Yes)P(CP=High|HD=Yes)/ P(BP=High)P(CP=Yes)
= 0.85*0.49*1.34 / 0.85*0.49 + 0.2*0.51 = 0.42

xieguahu
Posts: 50
Posted 16:37 Aug 30, 2009 |
It doesnot make sense, the p should be less or equal to 1
p0941 wrote:
HelloWorld wrote:
p0941 wrote:
HelloWorld wrote:
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

 

Because BP and HD are conditionally dependent and they are both conditionally independent on CP.

CP is dependent on HD though..

Good point. I update part b here:

 

P(Hb=Y) = P(Hb|D=H)P(D=H) +  P(Hb|D=UnHealthy)(1-P(D=H))= 0.6875
P(Hb=N) = 1-0.6875 = 0.3125

P(CP=Y) = P(CP=Y|HD=Y,Hb=Y)P(HD=Y,Hb=Y)+
          P(CP=Y|HD=Y,Hb=N)P(HD=Y,Hb=N)+
          P(CP=Y|HD=N,Hb=Y)P(HD=N,Hb=Y)+
          P(CP=Y|HD=N,Hb=N)P(HD=N,Hb=N) = 0.51
P(CP=Y|HD=Y) = 0.8+0.6=1.4
P(HD=Yes) = 0.49, from Lecture example -1

P(HD=Y|CP=H) = P(CP=Y|HD=Y)P(HD=Y)/P(CP=Y) = 1.4*0.49 / 0.51 = 1.34


P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes)
= P(HD=Yes,BP=High,CP=Yes) / P(BP=H, CP=Y)
= P(BP=High|HD=Yes)P(HD=Yes)P(CP=High|HD=Yes)/ P(BP=High)P(CP=Yes)
= 0.85*0.49*1.34 / 0.85*0.49 + 0.2*0.51 = 0.42

 

p0941
Posts: 95
Posted 18:10 Aug 30, 2009 |
xieguahu wrote:
It doesnot make sense, the p should be less or equal to 1
p0941 wrote:
HelloWorld wrote:
p0941 wrote:
HelloWorld wrote:
p0941 wrote:

= P(BP=High|HD=Yes)P(HD=Yes)(CP=Yes) / P(BP=High)P(CP=Yes)

How do you get that part? I got this..

 

Because BP and HD are conditionally dependent and they are both conditionally independent on CP.

CP is dependent on HD though..

Good point. I update part b here:

 

P(Hb=Y) = P(Hb|D=H)P(D=H) +  P(Hb|D=UnHealthy)(1-P(D=H))= 0.6875
P(Hb=N) = 1-0.6875 = 0.3125

P(CP=Y) = P(CP=Y|HD=Y,Hb=Y)P(HD=Y,Hb=Y)+
          P(CP=Y|HD=Y,Hb=N)P(HD=Y,Hb=N)+
          P(CP=Y|HD=N,Hb=Y)P(HD=N,Hb=Y)+
          P(CP=Y|HD=N,Hb=N)P(HD=N,Hb=N) = 0.51
P(CP=Y|HD=Y) = 0.8+0.6=1.4
P(HD=Yes) = 0.49, from Lecture example -1

P(HD=Y|CP=H) = P(CP=Y|HD=Y)P(HD=Y)/P(CP=Y) = 1.4*0.49 / 0.51 = 1.34


P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes)
= P(HD=Yes,BP=High,CP=Yes) / P(BP=H, CP=Y)
= P(BP=High|HD=Yes)P(HD=Yes)P(CP=High|HD=Yes)/ P(BP=High)P(CP=Yes)
= 0.85*0.49*1.34 / 0.85*0.49 + 0.2*0.51 = 0.42

 

Yes, thank you.  Update:

 

P(Hb=Y) = P(Hb|D=H)P(D=H) +  P(Hb|D=UnHealthy)(1-P(D=H))= 0.6875
P(Hb=N) = 1-0.6875 = 0.3125

P(CP=Y) = P(CP=Y|HD=Y,Hb=Y)P(HD=Y,Hb=Y)+
          P(CP=Y|HD=Y,Hb=N)P(HD=Y,Hb=N)+
          P(CP=Y|HD=N,Hb=Y)P(HD=N,Hb=Y)+
          P(CP=Y|HD=N,Hb=N)P(HD=N,Hb=N) = 0.51
P(CP=Y|HD=Y) = P(CP=Y|HD=Y,Hb=Y)P(HD=Y)P(Hb=Y)+P(CP=Y|HD=Y,Hb=N)P(HD=Y)P(Hb=N)= 0.8*0.49*0.6875+0.6*0.49*0.3125= 0.36
P(HD=Yes) = 0.49, from Lecture example -1

P(HD=Y|CP=H) = P(CP=Y|HD=Y)P(HD=Y)/P(CP=Y) = 0.36*0.49 / 0.51 = 0.34


P(HeartDisease=Yes|BloodPresure=High,ChestPain=Yes)
= P(HD=Yes,BP=High,CP=Yes) / P(BP=H, CP=Y)
= P(BP=High|HD=Yes)P(HD=Yes)P(CP=High|HD=Yes)/ P(BP=High)P(CP=Yes)
= 0.85*0.49*0.34 / 0.85*0.49 + 0.2*0.51 = 0.18

 

 

 

Pallavi0205
Posts: 30
Posted 18:50 Aug 30, 2009 |

Here the solution for 16. i got 16a same as abhishek and grady but solution b is different.

Attachments:
Pallavi0205
Posts: 30
Posted 19:22 Aug 30, 2009 |

made one calculation error in final step. I just fixed it.

Attachments:
HelloWorld
Posts: 88
Posted 20:08 Aug 30, 2009 |

I thought of this for part B

 

(b)

P(HD=Yes | BP= High, CP=Yes)

= P(HD=Yes) = 0.49

The probability of Heart Disease (HD) is independent of the Blood Pressure (BP) and Chest Pain (CP), therefore, we can just get rid of them. From the diagram, we could see that Heart Disease is the one that determines the probability of Blood Pressure and Chest Pain, not the other way around.

 

Last edited by HelloWorld at 20:08 Aug 30, 2009.
pratyush
Posts: 6
Posted 20:17 Aug 30, 2009 |
Pallavi0205 wrote:

made one calculation error in final step. I just fixed it.

 

P (CP= Y|HD = Y) =

P (CP = Y/HD = Y, Hb= Y) P (HD = Y) P (Hb = Y) + P (CP = Y/HD =Y, Hb =N) P (HD = Y) P (Hb = N)

= 0.8 x 0.49 x 0.6875 + 0.6 x 0.51 x 0.3125 = 0.365 (marked as red :Do you think it should be P (HD=N) since u considered its value as 0.51) Please let me know.

 

 

Last edited by pratyush at 20:18 Aug 30, 2009.
Pallavi0205
Posts: 30
Posted 20:24 Aug 30, 2009 |

No I had made mistake while taking values. It's HD=Y so it's 0.49.

Attachments:
batcat
Posts: 11
Posted 20:50 Aug 30, 2009 |

For part a) i hope xieguahua's solution is correct (answer .34)

3 people got solution with .085 but final answer is illogical to me

We are given that if if D = Healthy and E = Yes, then prob of HD is .25

and if D = Healthy and E = No, then prob is .55 .

But then if we only know D = Healthy, prob of HD is .085??

Last edited by batcat at 20:52 Aug 30, 2009.
HelloWorld
Posts: 88
Posted 20:51 Aug 30, 2009 |
batcat wrote:

For part a) i hope xieguahua's solution is correct

3 people got solution with .085 but final answer is illogical to me

We are given that if if D = Healthy and E = Yes, then prob of HD is .25

and if D = Healthy and E = No, then prob is .55 .

But then if we only know D = Healthy, prob of HD is .085??

Exactly! that's what I just thought after having discussion with him

HelloWorld
Posts: 88
Posted 21:17 Aug 30, 2009 |
Pallavi0205 wrote:

No I had made mistake while taking values. It's HD=Y so it's 0.49.

I think this is the problem:

 

P (CP= Y|HD = Y) =

P (CP = Y/HD = Y, Hb= Y) P (HD = Y) P (Hb = Y) + P (CP = Y/HD =Y, Hb =N) P (HD = Y) P (Hb = N) 

= 0.8 x 0.49 x 0.6875 + 0.6 x 0.49 x 0.3125 = 0.361

if you pay attention.. the underline part, what's the difference then if we're asking for P(CP=Y)? that's how you find P(CP=Y) right?

Check my solution.. I agree with Guanghua's solution for part a

 

Attachments:
Last edited by HelloWorld at 21:19 Aug 30, 2009.
HelloWorld
Posts: 88
Posted 21:51 Aug 30, 2009 |
HelloWorld wrote:
Pallavi0205 wrote:

No I had made mistake while taking values. It's HD=Y so it's 0.49.

I think this is the problem:

 

P (CP= Y|HD = Y) =

P (CP = Y/HD = Y, Hb= Y) P (HD = Y) P (Hb = Y) + P (CP = Y/HD =Y, Hb =N) P (HD = Y) P (Hb = N) 

= 0.8 x 0.49 x 0.6875 + 0.6 x 0.49 x 0.3125 = 0.361

if you pay attention.. the underline part, what's the difference then if we're asking for P(CP=Y)? that's how you find P(CP=Y) right?

Check my solution.. I agree with Guanghua's solution for part a

I take that back, i didn't pay closer attention.. it's different.. so don't worry about my comment above, here's my updated solution, the difference is I calculated P(CP=Yes | HD=Yes) without having to multiply it again with P(HD=Yes) (i.e. in agreement with part A of GX's solution)

Attachments:
abhishek_sharma
Posts: 79
Posted 22:59 Aug 30, 2009 |
HelloWorld wrote:
batcat wrote:

For part a) i hope xieguahua's solution is correct

3 people got solution with .085 but final answer is illogical to me

We are given that if if D = Healthy and E = Yes, then prob of HD is .25

and if D = Healthy and E = No, then prob is .55 .

But then if we only know D = Healthy, prob of HD is .085??

Exactly! that's what I just thought after having discussion with him.

Well, its obvious if you know ONLY D=healthy then prob of HD is less compared to D=healthy , E=yes/No where an extra parameter is known in addition to D=healthy. I hope I am able to clearify. Still if you find any mistake kindly let me know. Thanks.

cysun
Posts: 2935
Posted 16:03 Aug 31, 2009 |

Guanhua's solution (3rd post from the top) is correct. Note that the last line of p0941's solution should be "= 0.85*0.49 / (0.85*0.49 + 0.2*0.51) = 0.8" instead of "= 0.85*0.49 / 0.85*0.49 + 0.2*0.51 = 0.8".