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# My uphill battle with Statistics watch

1. I'm your typical engineering student that's done only mechanics modules before and while now at uni, this is my first experience with statistics.

I've been trying to get better but not hard enough so I'm making this thread to keep a public record of my progress.

I'll also be asking question (not for full solutions this time ) but I'll post a question and may ask for a numerical answer to check my work to I'll post photos to ask if they're correct

And don't be shy other people, you can ask your statistics inquiries here as well, we can all learn together

Happy Posting!

Zacken (tag others that may be interested)
2. Remember.

AND rule is multiplication
3. Remember.

Hakuna /=/ Matata
4. Sounds good, I'll be happy to help with what I can.
5. Man where was this help when i was failing 2 years ago lol
6. (Original post by MevMev)
Man where was this help when i was failing 2 years ago lol
Right here, the maths forum has been going on for a decade or so.
7. (Original post by Zacken)
Sounds good, I'll be happy to help with what I can.

For part a, I got 0.1 but they say the answer is 0.28?

I tried using the conditional probability P(IC | CP) = P(IC ∩ CP) / P(CP) = (0.7 x 0.1)/07 = 0.1

Note: IC - internal corrosion, CP - corrosion present , CF = corrosion falsely detected

My issues
1) Not entirely sure how to get the intersection part in the formula
2) I don't understand partitioning events with its compliment.
8. (Original post by ServantOfMorgoth)

For part a, I got 0.1 but they say the answer is 0.28?

I tried using the conditional probability P(IC | CP) = P(IC ∩ CP) / P(CP) = (0.7 x 0.1)/07 = 0.1

Note: IC - internal corrosion, CP - corrosion present , CF = corrosion falsely detected

My issues
1) Not entirely sure how to get the intersection part in the formula
2) I don't understand partitioning events with its compliment.
You're getting confused on what is conditioned on what. So let D be the event that the test detects corrosion and let CP be the event "corrosion present". Then you are given P(D | CP) = 0.7 and that P(D | ~CP) = 0.2 and P(CP) = 0.1.

You are asked to work out P(CP | D). Can you do that via Bayes' theorem? That is

9. (Original post by Gregorius)
You're getting confused on what is conditioned on what. So let D be the event that the test detects corrosion and let CP be the event "corrosion present". Then you are given P(D | CP) = 0.7 and that P(D | ~CP) = 0.2 and P(CP) = 0.1.

You are asked to work out P(CP | D). Can you do that via Bayes' theorem? That is

What does the sum of the denominator symbolize? Like I only know Bayes theorem to be
10. (Original post by Zacken)
Right here, the maths forum has been going on for a decade or so.
What if you post too much surely you'd get annoyed
11. (Original post by ServantOfMorgoth)
What does the sum of the denominator symbolize? Like I only know Bayes theorem to be
The sum in the denominator I have given is a special case of the one you have there. You have as a partition of the the probability space ; I have and not partitioning .
12. (Original post by Gregorius)
The sum in the denominator I have given is a special case of the one you have there. You have as a partition of the the probability space ; I have and not partitioning .
Oh ok, I think I understand but in part b I think I need to find P(CP|~D) ? I'm not sure how to operate with ~D since they didn't give D separately?
13. (Original post by Samistrawberry)
What if you post too much surely you'd get annoyed
Surely they would over time but there's lots of people to help
14. (Original post by Samistrawberry)
What if you post too much surely you'd get annoyed
That would defeat the purpose of the forum, everybody is welcome to post as many question as they'd like. I've helped a fair few people on here, and amongst those who post repeated questions, I can never once say that I felt annoyed by them as long as they were courteous.
15. (Original post by ServantOfMorgoth)
Oh ok, I think I understand but in part b I think I need to find P(CP|~D) ? I'm not sure how to operate with ~D since they didn't give D separately?
What you can work out from what they have given you is P(~D | CP) = 1 - P(D | CP). You can now use Bayes to invert that one.
16. (Original post by ServantOfMorgoth)
Surely they would over time but there's lots of people to help
Haha I've stopped studying maths but they're a lot more fun to help solve then chem questions.

(Original post by Zacken)
That would defeat the purpose of the forum, everybody is welcome to post as many question as they'd like. I've helped a fair few people on here, and amongst those who post repeated questions, I can never once say that I felt annoyed by them as long as they were courteous.
Sometimes I'm just too dumb to understand and it might frustrate others.
17. (Original post by Samistrawberry)
Sometimes I'm just too dumb to understand and it might frustrate others.
Give it a try next time you get stuck, you'll be pleasantly surprised.
18. I'll be back on Wednesday afternoon as per my schedule permits

Thanks all so far Gregorius
19. (Original post by Zacken)
Give it a try next time you get stuck, you'll be pleasantly surprised.
Chemistry is a slight bit more confusing to understand/explain. I often just shake my head in class saying I understand when I don't when miss explains.

Will try to make use of the Chem forum and try not to spend all my time on the Chat section
20. Guys I flopped my statistics incourse test

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