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Section I: Multiple Choice (1 Viewer)

jer3m

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No. If there are 31 days in October and there is an expected 8 rainy days and then the first day is rainy and they expect the second to be it becomes 8/31 x 7 (the remaining expected) x 30 (the remaining days). That's how in THEORY probability works. If you threw a coin 9 times and it was heads all the time the final chance would still be 1/2 for a tail. This question however doesn't follow that idea.
It is 8/31 x 7/30 hence A.
 

jer3m

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If the question did not say consecutive then yes you would be right. But it is consecutive. Thereby you have to remove the previous day/rain. Think of it as a tree diagram
 

ag6858

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Haha I figure I'm not going to be able to change your mind on this. Let's just agree to disagree
 

Hanakalala

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1. D
2. B
3. C
4. B
5. B
6. D
7. D
8. A
9. C
10. A
11. C A
12. C
13. B
14. D B
15. B
16. A
17. B
18. D
19. A
20. D B
21. A C
22. B
23. C
24. D
25. A

^ Thoughts?
Can't remember exactly all mine (forgot to circle) but the ones i recognised i put different are in bold... i could be wrong so don't panic too much haha
In the test, I got the same as what you just said besides question 20. I circled D, although I think I am wrong.. Anyone care to shed some light on the working out for Question 20?
 

ag6858

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No. If there are 31 days in October and there is an expected 8 rainy days and then the first day is rainy and they expect the second to be it becomes 8/31 x 7 (the remaining expected) x 30 (the remaining days). That's how in THEORY probability works. If you threw a coin 9 times and it was heads all the time the final chance would still be 1/2 for a tail. This question however doesn't follow that idea.
It is 8/31 x 7/30 hence A.
http://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-mg-maths-general-2.pdf
Or not...
 

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