Who wants to do some easy STAT?

[Syringer]

I'm braindead right now, so any help would be greaatt..

Girl has 3 exams that she will take in order, if she fails any exam she's not allowed to go on. Prob that she passes the first exam is .9. If she passes the 1st exam, then the conditional probability that she passes the second one is .8, and if she passes both the first and 2nd exam, then the conditional prob that she passes on the 3rd exam is .7.

a) What is the prob that she passes all three?

Do you just multiply all of them together? That seems like it'd be too easy..

b) Given that she did not pass all 3, what is the conditional prob that she failed the 2nd exam?

 

[r6ashih]

A.) spyder
B.) Spider


1. yes you multiply them together
2. this sounds like a trick question. .. given that she did not pass all 3 = 100% failure?
if its not that.. youll have to make one of these < trees mapping out the possibilities with their %.


what stats class r u in? stats 11?

 

[Syringer]

You go to UCLA too?

I'm in 100b..which is pathetic that I still can't do this crap.

 

[quakefiend420]

i have absolutely nothing useful to add to this thread except...






























thought i was gonna say spider didnt you

 

[r6ashih]

yeah i go to ucla.. but i'm only in stats 10

 

[r6ashih]

i updated my response to the second part..

 

[chuckywang]

Originally posted by: Syringer
I'm braindead right now, so any help would be greaatt..

Girl has 3 exams that she will take in order, if she fails any exam she's not allowed to go on. Prob that she passes the first exam is .9. If she passes the 1st exam, then the conditional probability that she passes the second one is .8, and if she passes both the first and 2nd exam, then the conditional prob that she passes on the 3rd exam is .7.

a) What is the prob that she passes all three?

Do you just multiply all of them together? That seems like it'd be too easy..

b) Given that she did not pass all 3, what is the conditional prob that she failed the 2nd exam?

a) Yes, just multiply them all.

b) Prob she did not pass = 1-answer in part a. Let this be equal to X.
Prob. that she failed the second = 0.9*0.2. Let this be equal to Y.

Conditional Probability that she failed 2nd exam = Y/X.

 

[Syringer]

How is it 1-answer in part a? Wouldn't it p(failed exam 1) + p(passed 1, failed 2) + p(passed 1 and 2, failed 3)?

 

[jst0ney]

I barely passed stats in undergrad. I hated my teach. It made grad school hard but oh well what can you do?













spider - is this getting old yet

 

[chuckywang]

Originally posted by: Syringer
How is it 1-answer in part a? Wouldn't it p(failed exam 1) + p(passed 1, failed 2) + p(passed 1 and 2, failed 3)?

Compute both of them out. You'll find they are one and the same. Amazing how math works.

 

[Syringer]

Ah, amazing