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November 08, 2025, 04:44:06 am
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Author Topic: bye  (Read 1589 times)  Share 

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uranrahabatula

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bye
« on: September 24, 2009, 12:30:02 am »
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bye
« Last Edit: November 29, 2009, 12:13:15 am by uranrahabatula »
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n.f

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Re: Umat problems from section 1
« Reply #1 on: September 24, 2009, 11:52:27 am »
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Q42. If it is known that 7-30 only have 3/5 prizes then 1-6 must contain 2/5 prizes. So a 2/6 chance, simplify to 1/3
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n.f

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Re: Umat problems from section 1
« Reply #2 on: September 24, 2009, 07:00:24 pm »
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Q43. It is 3/5. If there's a 1/6 chance the first box has one of three prizes and finds it does not then the second box will have a 3/5 chance of having a prize.
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QuantumJG

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Re: Umat problems from section 1
« Reply #3 on: September 24, 2009, 07:05:05 pm »
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Looking through the UMAT papers you could actually learn how to do self diagnosis. lol.

Looks like an intense test.
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2008: Finished VCE

2009 - 2011: Bachelor of Science (Mathematical Physics)

2012 - 2014: Master of Science (Applied Mathematics/Mathematical Physics)

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Semester 1:[/b] Engineering Mechanics, Fluid Mechanics, Engineering Risk Analysis, Sustainable Infrastructure Engineering

Semester 2:[/b] Earth Processes for Engineering, Engineering Materials, Structural Theory and Design, Systems Modelling and Design

n.f

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Re: Umat problems from section 1
« Reply #4 on: September 24, 2009, 07:15:43 pm »
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Sorry can't work out 44 atm.
Half of the time he's wrong of his 24 'known'. He says 3/5 prizes are in these 24. The chance of one of box 2 (of 6 unknown) has 1/5 prizes is...
[Above is my train of thought for when I look later]
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