probability question (1 Viewer)

[paper cup]

Braindead, braindead, braindead....I can't believe I can't do this question, it's 2 Unit, and probability, the topic that you will always obtain full marks in unless you read the question wrong. *bites nails*

Help, anyone?

A manufacturer finds that 15% of the articles made are defective. If three articles are taken at random, find the probability that there are more defectives than non defectives.

So P(D>nD) = P(nDDD) + P(DnDD) + P(DDnD) + P(DDD) = (3/20 x 3/20 x 17/20) cubed + 0.00375 = something very different to the given answer of 0.0225 (= 3/20 x 3/20).

What the fuk am I missing?

 

[withoutaface]

3C2(.15)²(.85)+(.15)³

 

[paper cup]

I know, but that's binomial probability - 3 unit stuff, how would you use 2 unit probability to solve it? and by the way it's 3 x what's inside the brackets, not cubed.

 

[sammeh]

you wouldnt. u can use any of the 3 unit only stuff to answer 2 unit questions. same goes for 4 unit to answer 3 unit, etc etc. hell, u could probly break out some extra-curricular stuff if u really felt it necessary. hsc markers will go out of their way to validate your work, they're trying to give you the HIGHEST mark possible.

dont waste your time trying to double up your study. just stick with the more sensible 3 unit stuff ^^

 

[pc_wizz]

you wouldnt. u can use any of the 3 unit only stuff to answer 2 unit questions. same goes for 4 unit to answer 3 unit, etc etc. hell, u could probly break out some extra-curricular stuff if u really felt it necessary. hsc markers will go out of their way to validate your work, they're trying to give you the HIGHEST mark possible.

dont waste your time trying to double up your study. just stick with the more sensible 3 unit stuff ^^

u sure ...

our teacher, and my tutor have told me repeatedly to stick to 2u formula in 2u, 3u in 3u, 4u in 4u etc ... even though it makes it easier say using 3u applications in 2u, or 4u in 3u, but i dont want to risk the possibility of not being awarded set criteria marking ...

 

[CrashOveride]

I know, but that's binomial probability - 3 unit stuff, how would you use 2 unit probability to solve it? and by the way it's 3 x what's inside the brackets, not cubed.

For the component of probability that all are defective, we have 0.15 x 0.15 x 0.15 i.e. (0.15)³.

As for boundaries between the various streams of maths, it's all just maths.

 

[ngai]

alright then, stick to the 2u stuff, and case bash the question hahaha
case 1: all 3 defective, (0.15)^3
case 2: 1st two are defective, ...
case 3: last two are defective, ...
case 4: 1st and last are defective, ...
add them up

kinda stupid to have to stick to 2u tho...maths is maths..if wat u do is right, then they should give u ur marks, and i would argue with them until they do

 

[paper cup]

oh! So I was missing the last/first defective combo..thank you.

But what about the poor misguided souls who do only 2 unit maths, how would they go about doing that question? It always is best to stick to whatever 'unit' formulaes...

 

[CrashOveride]

What withoutaface used up there was just logic. Anyone with appreciation of logic could see how that works up there.

 

[ngai]

But what about the poor misguided souls who do only 2 unit maths, how would they go about doing that question? It always is best to stick to whatever 'unit' formulaes...

few things u can do:
1) learn more maths, outside the 2u sillybus
2) case bash the question like wat i wrote
3) and for some other questions, count them all individually

 

[withoutaface]

Another possible solution, an idea from which binomial probability is derived, is:

If all 3 are defective then (.15)³
If the first is working .85*.15*.15
""""second""" .15*.85*.15
" " " third " " " .15*.15*.85

Add these together and you get your answer

EDIT: Whoops, Ngai already did that

 

[jhakka]

Just use a probability tree.

 

[~*HSC 4 life*~]

Just use a probability tree.

i hate those, mine always look wonked

 

[Jago]

Just use a probability tree.

yeah, can't go wrong there.