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March 29, 2024, 12:13:02 am

Author Topic: 3U Maths Question Thread  (Read 1230245 times)  Share 

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Neutron

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Re: 3U Maths Question Thread
« Reply #495 on: August 07, 2016, 07:12:08 pm »
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It's from one of my school's past papers :/

RuiAce

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Re: 3U Maths Question Thread
« Reply #496 on: August 07, 2016, 07:15:08 pm »
+1
Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one:

In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by $250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed $5 000 000 when it is finally won.

The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D:

Thank you legends!!



I was about to ask where the source of the question was. Can you ask your school for some clarification?

Extra reading: Statisticians refer to these independent events as following a "geometric" distribution. A geometric distribution measures the probability of the first success after having (n-1) fails. It follows the "negative-binomial" distribution for n=1.

The probability mass function for the geometric distribution is (1-p)n-1p, where p is the probability of success.
« Last Edit: August 07, 2016, 07:20:56 pm by RuiAce »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #497 on: August 07, 2016, 07:41:57 pm »
+1
It's from one of my school's past papers :/

Yep, consenting with Rui above, his method is how I would solve it. They've solved for the probability that the lottery is won as soon as the prize exceeds 5 million, but as you correctly say, it could be won when the prize is $100 million ;D

massive

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Re: 3U Maths Question Thread
« Reply #498 on: August 07, 2016, 08:38:02 pm »
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yoo guys, howdu do part ii?

RuiAce

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Re: 3U Maths Question Thread
« Reply #499 on: August 07, 2016, 09:25:51 pm »
+1

Neutron

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Re: 3U Maths Question Thread
« Reply #500 on: August 07, 2016, 10:39:17 pm »
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Ahh thank you so so much!! One more thing from me (sorry) :o This is the question:

The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB?

How do you know whether it's 3:10 or 10:3? :O Thank you!

RuiAce

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Re: 3U Maths Question Thread
« Reply #501 on: August 07, 2016, 10:41:34 pm »
+1
Ahh thank you so so much!! One more thing from me (sorry) :o This is the question:

The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB?

How do you know whether it's 3:10 or 10:3? :O Thank you!
Match it up:

A---P-------B
  3  :     7

|-----10------|

So A is closer to P than it is to B. So since P comes first:
So 3:10

Jakeybaby

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Re: 3U Maths Question Thread
« Reply #502 on: August 07, 2016, 10:49:12 pm »
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Ahh thank you so so much!! One more thing from me (sorry) :o This is the question:

The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB?

How do you know whether it's 3:10 or 10:3? :O Thank you!
Spoiler

(see spoiler) Therefore, following the 2 steps, the external ratio which A divides PB is 3:10

Edit: Added Spoiler
2016 ATAR: 98.60

2020: Bachelor of Finance @ University of Adelaide

Recipient of the 2017 University of Adelaide Principals' Scholarship

Neutron

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Re: 3U Maths Question Thread
« Reply #503 on: August 07, 2016, 11:07:16 pm »
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Match it up:

A---P-------B
  3  :     7

|-----10------|

So A is closer to P than it is to B. So since P comes first:
So 3:10

But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question:

Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective?

Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said:

P(exactly one computer defective)
=P( 1 defective from A, 0 from B) + P(0 from A, 1 from B)
3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095

Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ )

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #504 on: August 07, 2016, 11:19:50 pm »
+1
But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question:

Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective?

Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said:

P(exactly one computer defective)
=P( 1 defective from A, 0 from B) + P(0 from A, 1 from B)
3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095

Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ )

Did this in my 3U lecture ;)

So there is a slight difference here. Warehouse A has a probability of 2% that it is defective for every computer. Warehouse B has two defective computers. So, the probability is dependent on how many computers are left in the warehouse! As you buy computers, the probability turns from 2/100, to 2/99, 2/98, etc. The binomial probability idea can only be applied when the probability is identical with every repetition of the experiment, called a Bernoulli Trial.

The multiplying by three? That comes from the fact that the defective computer can be the first computer bought from B, OR the second, OR the third. We add the probability of all three possibilities, but that ends up being the same as multiplying by three ;D

RuiAce

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Re: 3U Maths Question Thread
« Reply #505 on: August 07, 2016, 11:31:10 pm »
+1
But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question:

Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective?

Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said:

P(exactly one computer defective)
=P( 1 defective from A, 0 from B) + P(0 from A, 1 from B)
3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095

Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ )
You do? I always just matched up the ordering.

If it were division of BP then I would've gotten 10:3
Did this in my 3U lecture ;)

I thought the question looked familiar..

conic curve

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Re: 3U Maths Question Thread
« Reply #506 on: August 08, 2016, 10:56:57 am »
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Find domain and range of the following functions and sketch their graphs:

1. y=-3sin^-1 x

For domain I got -1≤x≤1

For range I was not too sure???

2. y=cos^-1 2x

For domain, I got -1/2≤x≤1/2

Range I'm not too sure. Is it 0≤cos^-1 2x≤π?

3. y=3tan^-1 (x/2)

Domain: all x
Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong)

Point of inflection: let x/2=0
x=0
Sub in LHS point (and I'm not sure of what I need to do)

Could someone here please help me

Thanks

massive

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Re: 3U Maths Question Thread
« Reply #507 on: August 08, 2016, 11:11:55 am »
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hey, how do you do this guys?

massive

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Re: 3U Maths Question Thread
« Reply #508 on: August 08, 2016, 11:20:36 am »
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Find domain and range of the following functions and sketch their graphs:

1. y=-3sin^-1 x

For domain I got -1≤x≤1

For range I was not too sure???

The range for this is -3pi≤y≤0

Find domain and range of the following functions and sketch their graphs:

2. y=cos^-1 2x

For domain, I got -1/2≤x≤1/2

Range I'm not too sure. Is it 0≤cos^-1 2x≤π?

The range for this is just the normal 0≤y≤pi

Find domain and range of the following functions and sketch their graphs:

3. y=3tan^-1 (x/2)

Domain: all x
Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong)

Point of inflection: let x/2=0
x=0
Sub in LHS point (and I'm not sure of what I need to do)

The range for this is -3pi/2<y<3pi/2
im not really sure how you get point of inflexion (sorry) but im pretty sure its still at the origin for this.

RuiAce

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Re: 3U Maths Question Thread
« Reply #509 on: August 08, 2016, 11:44:40 am »
+1
Find domain and range of the following functions and sketch their graphs:

1. y=-3sin^-1 x

For domain I got -1≤x≤1

For range I was not too sure???

2. y=cos^-1 2x

For domain, I got -1/2≤x≤1/2

Range I'm not too sure. Is it 0≤cos^-1 2x≤π?

3. y=3tan^-1 (x/2)

Domain: all x
Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong)

Point of inflection: let x/2=0
x=0
Sub in LHS point (and I'm not sure of what I need to do)

Could someone here please help me

Thanks