If you want any help with maths, ask me here

pinkbear

Oops, sorry yes, I see it now, and agree with those answers. I should probably revise Picking and Choosing to avoid me making mistakes like this!

(ii) So the way I did it above shows you the probability of choosing them in that particular order (i.e. P, not C). Now you need to see how many different ways can the 4 disks be ordered: i.e. 4x3x2x1=24. I'm not looking at the order that they are in (i.e. I know that Y G Blue Black is the same as Blue Y G Black etc.), but looking at the chances I could have chosen them. I.e. I could have chosen Yellow followed by Green followed by Blue followed by Black, so those probabilities have to be added. If there are 24 different possibilities that have to be added, it's the same as multiplying by 24.
So multiply these: 3/728 x 24 = 9/91

(iii) Similarly the answer I got was for Black, Black, Not Black, Not black.
If you think about the order of these again, there are 6 possible orders these could have been chosen.
(BBxx, BxxB, BxBx, xBBx, xBxB, xxBB). So adding the probability of each is the same as multiplying by 6.
So multiply these: 1/120 x 6 = 1/20

That make sense?

thetalker

Wow that actually makes a lot of sense but I'm a bit surprised it works

You see for ii, you found the amount of ways 4 could be arranged and multiplied but I would've imagined you couldn't do that because for instance.

(5/16 * 3/15 * 6/14 * 2/13) = 3/728
rearranged would give
(3/16 * 5/15 * 2/14 * 6/13) = 3/728

Oh wait....it is the same. I didn't expect that. It seems strange. Is there some simple logic here I'm failing to realize?

But apart from that I get it, its actually something I prefer over P and C

pinkbear

The reason that you get 3/728 both ways above is that they are both (tops by tops)/(bottoms by bottoms) so the order is unimportant.

Think of it simpler, e.g. having 2 green and 2 black. What is the probability of choosing one of each?
You could choose green followed by black, or black followed by green, and both of those would give you one of each.
The chances of choosing green is 1/2, chances of choosing black is 1/2, chances of choosing both if order mattered is 1/2 x 1/2 = 1/4.
However if you look at the number of ways that green and black could be chosen it's 2 - i.e. GB or BG. So multiply 1/4 by 2 and you get 1/2 which is your answer.

thetalker

Oh I see, so if it was addition it wouldn't be possible.
Btw I remember you mentioned using math in your job, would you be a physicist or engineer? Because I've been wondering about a career in those fields.

pinkbear

thetalker wrote: »

Oh I see, so if it was addition it wouldn't be possible.
Btw I remember you mentioned using math in your job, would you be a physicist or engineer? Because I've been wondering about a career in those fields.

Don't know what you mean when you say "if it was addition it wouldn't be possible". Explain further, and I'll try to answer.

Sent you a private message regarding what I work at.

thetalker

pinkbear wrote: »

Don't know what you mean when you say "if it was addition it wouldn't be possible". Explain further, and I'll try to answer.

Sent you a private message regarding what I work at.

Oh its nothing really, I just was pointing out how if we were adding (6/16 + 5/14) we wouldnt be able to claim its the same as (5/16 + 6/14) for obvious reasons.

Take Your Pants Off

I know this one's quite a lot, but I don't even know where to start in these.
It's 12-14. And for 10 part 2, I make my sum out to be like that, 5/14 × 3/8 × 2/7 but i get 15/396. Any help would be appreciated.
Thank you

pinkbear

Take Your Pants Off wrote: »

I know this one's quite a lot, but I don't even know where to start in these.
It's 12-14. And for 10 part 2, I make my sum out to be like that, 5/14 × 3/8 × 2/7 but i get 15/396. Any help would be appreciated.
Thank you

Answers to 12-14(a) attached. I don't have time to look at 10 or 14(b) tonight but I should do tomorrow night. Write back if you don't follow what I've done for the first 3.

Take Your Pants Off

pinkbear wrote: »

Answers to 12-14(a) attached. I don't have time to look at 10 or 14(b) tonight but I should do tomorrow night. Write back if you don't follow what I've done for the first 3.

Thank you so much. Always helpful in this pinkbear!.
Ill try it out tomorrow and see if I can work with the answers.

platypus

Surely easier do 14a by using 1 - P(none same)
So 1-(1*364/365*363/365*362/365)

Ie 1st can be born any day
2nd on 364 other days
3rd on 363 other days ( different than 1st 2)
4th on 362 other days ( different than other 3)

14b can be approached in a similar fashion

pinkbear

Yes, platypus is right, it can be done that way either.

So 14(ii) using that method: 1-P(all have different birthdays) = 1-364/365 x 364/365 x …. x 341/365 = 1 - .458 = 0.532

Back to 10(ii)

P(Person 1 chooses 2 consonants) x P(person 2 choses 2 vowels)
P(C)P(C) x P(V)P(V)
3/8 x 2/7 x 5/8 x 4/7 = 15/392

Hope those are right.

platypus

pinkbear wrote: »

Yes, platypus is right, it can be done that way either.

So 14(ii) using that method: 1-P(all have different birthdays) = 1-364/365 x 364/365 x …. x 341/365 = 1 - .458 = 0.532

Back to 10(ii)

P(Person 1 chooses 2 consonants) x P(person 2 choses 2 vowels)
P(C)P(C) x P(V)P(V)
3/8 x 2/7 x 5/8 x 4/7 = 15/392

Hope those are right.

I'd say need to double that answer in 10(ii), can be done in opposite order is 1st picks vowels 2nd picks consonants

pinkbear

Yes, you're right platypus, so it would be double 15/392, which is 15/196. That correct platypus?

Take Your Pants Off

pinkbear wrote: »

Yes, platypus is right, it can be done that way either.

So 14(ii) using that method: 1-P(all have different birthdays) = 1-364/365 x 364/365 x …. x 341/365 = 1 - .458 = 0.532

Back to 10(ii)

P(Person 1 chooses 2 consonants) x P(person 2 choses 2 vowels)
P(C)P(C) x P(V)P(V)
3/8 x 2/7 x 5/8 x 4/7 = 15/392

Hope those are right.

Would the atleast method work for when they are not replaced, i.e. in this question(1.iv)
I tried 1-1/16=15/16
Then 15/16 * 14/16( since theyr not replaced) and then i minus by 1 and mulitply by two since it can be done twice and i dont get the answer which is at the back(11/14)

pinkbear

Take Your Pants Off wrote: »

Would the atleast method work for when they are not replaced, i.e. in this question(1.iv)
I tried 1-1/16=15/16
Then 15/16 * 14/16( since theyr not replaced) and then i minus by 1 and mulitply by two since it can be done twice and i dont get the answer which is at the back(11/14)

Q. A bag contains eight black, three white, and five red beads. Three beads are picked at random without replacement. Find the probability that at least two are the same colour.

A. This can happen 3 ways….
At least 2 black
At least 2 white
At least 2 red

P(2 black): 8/16 x 7/15 = 7/30
P(2 white): 3/16 x 2/15 = 1/40
P(2 red): 5/16 x 4/15 = 1/12

P(2 black) + P(2 white) + P(2 red) = 7/30 + 1/40 + 1/12 = 41/120 = 0.342

Mr Rhode Island Red

Hi, was wondering if you know what to do with the following (I'm writing out in words because I'm on mobile)

Find the limit, when n is approaching infinity, of n^2 + 9 divided by 2n^2 +9n

Liordi

Could be wrong but I thought it would just be 0 because anything divided by infinity is 0.

pinkbear

You have to divide everything by the highest power of n, i.e. n^2. So it's (1+9/n^2)/(2+9/n). As n approaches infinity, the two 9/ terms become zero, so the answer is 1/2. Generally if you put a big value into your calculator for n, say n=10000, you should get a good idea of what the answer should be.

Liordi

pinkbear wrote: »

You have to divide everything by the highest power of n, i.e. n^2. So it's (1+9/n^2)/(2+9/n). As n approaches infinity, the two 9/ terms become zero, so the answer is 1/2. Generally if you put a big value into your calculator for n, say n=10000, you should get a good idea of what the answer should be.

Do you need to do this in all questions that are like this?

Mr Rhode Island Red

pinkbear wrote: »

You have to divide everything by the highest power of n, i.e. n^2. So it's (1+9/n^2)/(2+9/n). As n approaches infinity, the two 9/ terms become zero, so the answer is 1/2. Generally if you put a big value into your calculator for n, say n=10000, you should get a good idea of what the answer should be.

Massive thanks for the quick response

pinkbear

Liordi wrote: »

Do you need to do this in all questions that are like this?

I've done examples (attached) of all the kinds I can think of. I don't think examples like E.g. 6 has ever come up, but no harm to see it anyway.

platypus

pinkbear wrote: »

Q. A bag contains eight black, three white, and five red beads. Three beads are picked at random without replacement. Find the probability that at least two are the same colour.

A. This can happen 3 ways….
At least 2 black
At least 2 white
At least 2 red

P(2 black): 8/16 x 7/15 = 7/30
P(2 white): 3/16 x 2/15 = 1/40
P(2 red): 5/16 x 4/15 = 1/12

P(2 black) + P(2 white) + P(2 red) = 7/30 + 1/40 + 1/12 = 41/120 = 0.342

If you do the question that way you need to take account of the last bead being not the same for each 2 the same option and the orders they can come in. So P(2black) should be 8/16 X 7/15 X 8/14 x3(where 8/14 is probability 3rd bead not black and 3 is number of ways we can get 2 black one other)). We also need to add on the probabilities of all 3 beads being the same.

I prefer the approach of using 1-P(all different)

I would use combinations(A big advantage of using combinations is they take all possible orders into account).
8C1 is number of ways we can pick 1 black
3C1 is number of ways we can pick 1 white
5C1 is number of ways we can pick 1 red
16C3 is number of ways we can pick 3 objects from 16

So 1-((8C1X 3C1 X 5C1)/16C3)=11/14

Using conventional probability

1- (8/16 X 3/15 X 5/14 X 3!)

3! is the number of ways we can arrange 3 different objects(The top and bottomlines of fractions stay the same no matter which order so can just do one order and multiply by 3!

I'm sure there are other possible approaches. In ahurry now but will explain above further later if anyone wishes

sarakhan

hi pinkbear.
will i be able to get your email because i have higher level maths mock paper 2017 and i'm finding it very difficult and if i could send you pics of it and if you could help me with it. please.
thank you

pinkbear

Private message sent Sarakhan

Mr Rhode Island Red

I've a few real humdingers that have been causing me a bit of bother, wondering if you could give them a look when you get a chance. Alot of it is calculus, I never seem to know when I should or shouldn't be using the chain rule and such.

1. Find dy/dx of the Natural Log of the square root of 1 + x^2

2. Explain why a line with the equation y=2x + c where c is an element of the real numbers cannot be a tangent to the graph f(x) = x-1 + 1/x-1

3. Evaluate the integral from 0 to pi/2 of sin 3x cos 2x dx

I seem to keep getting zero and I don't know why. Apparently it's supposed to be 3/5 but I can't see what I'm doing wrong with it.

4. Complete the square of the following: -80x^2 +69,000x -9,300,000 to write it in the form r(x-h)^2 +k where k,r and h are constants

I know I'm supposed to half the x coefficient, square it, add it on, and then take it away, but I can't seem to get it into the required format. This particular question is from a sample paper so I guess the figures may not work possibly.

5. The first derivative of 2x lnx is 2(1+lnx) hence, find the indefinite integral of lnx dx

I know integrating is supposed to be the reverse of differentiating but after that I'm lost.

6. Differentiate sin^2 (3x-1) with respect to x

I've tried this 3 times and got cos^2 (9x-3), 1/2 - Cos(6x-2)/2, and -sin(6x-2)/2 all of which seem to be wrong.

7. The line y=x+c is a tangent to the graph f(x)=2/1-2x where x cannot equal a half. Find 2 possible values of c, where c is an element of the real numbers.

Don't know where to start with this.

8. Q(t) = Ae^-bt where a=2.920 and b=0.1

Show that Q(t) is a constant multiple of Q(t-1) for t being greater than or equal to 1

I'm not really sure what they're asking me here. I've never seen "constant multiple" being mentioned before.


9. Find the first and second derivatives of f(x)=e to the power of -0.5x^2

This is just me being useless at differentiating. I tried something with the chain rule but it's not working out. Apparently the second derivative is supposed to be (x^2 -1)e to the power of -0.5x^2 but I don't know how to get there.

Any and all advice would be massively appreciated, it's a lot of stuff but if you could even point me in the right direction it would be a massive help.

VG31

I did Q3. If I have the time later I'll have a loook at a few more.

https://s4.postimg.org/yrzetrerx/IMG_0621.jpg

pinkbear

Hi Mr Rhode Island Red,

I'll have a look at those when I get a chance, but it might take me a couple of days. The only one I can do without thinking is number 8, so here it is:

8. Q(t) = (2.920)e-(0.1)t
Q(t-1) = (2.920)e-(0.1)(t-1) = (2.920)e-(0.1)t + 0.1) = (2.920)e-(0.1)t (2.920)e 0.1
But (2.920)e 0.1 gives a constant number, value unimportant, just call it k
→ Q(t-1) = k(2.920)e-(0.1)t = k Q(t) → proven

PinkBear

Mr Rhode Island Red wrote: »

I've a few real humdingers that have been causing me a bit of bother, wondering if you could give them a look when you get a chance. Alot of it is calculus, I never seem to know when I should or shouldn't be using the chain rule and such.

1. Find dy/dx of the Natural Log of the square root of 1 + x^2

2. Explain why a line with the equation y=2x + c where c is an element of the real numbers cannot be a tangent to the graph f(x) = x-1 + 1/x-1

3. Evaluate the integral from 0 to pi/2 of sin 3x cos 2x dx

I seem to keep getting zero and I don't know why. Apparently it's supposed to be 3/5 but I can't see what I'm doing wrong with it.

4. Complete the square of the following: -80x^2 +69,000x -9,300,000 to write it in the form r(x-h)^2 +k where k,r and h are constants

I know I'm supposed to half the x coefficient, square it, add it on, and then take it away, but I can't seem to get it into the required format. This particular question is from a sample paper so I guess the figures may not work possibly.

5. The first derivative of 2x lnx is 2(1+lnx) hence, find the indefinite integral of lnx dx

I know integrating is supposed to be the reverse of differentiating but after that I'm lost.

6. Differentiate sin^2 (3x-1) with respect to x

I've tried this 3 times and got cos^2 (9x-3), 1/2 - Cos(6x-2)/2, and -sin(6x-2)/2 all of which seem to be wrong.

7. The line y=x+c is a tangent to the graph f(x)=2/1-2x where x cannot equal a half. Find 2 possible values of c, where c is an element of the real numbers.

Don't know where to start with this.

8. Q(t) = Ae^-bt where a=2.920 and b=0.1

Show that Q(t) is a constant multiple of Q(t-1) for t being greater than or equal to 1

I'm not really sure what they're asking me here. I've never seen "constant multiple" being mentioned before.


9. Find the first and second derivatives of f(x)=e to the power of -0.5x^2

This is just me being useless at differentiating. I tried something with the chain rule but it's not working out. Apparently the second derivative is supposed to be (x^2 -1)e to the power of -0.5x^2 but I don't know how to get there.

Any and all advice would be massively appreciated, it's a lot of stuff but if you could even point me in the right direction it would be a massive help.

pinkbear

Hi Mr Rhode Island Red,

I think all these answers are correct. Feel free to get back in touch if you need any more clarification.

PinkBear

Ponguin

Would anyone have a handy list of everything we need to derive/prove it would be really helpful! Last week our teacher surprised us saying that we needed to be able to derive the amortisation formula

Oathkeeper

Hi thanks for all you have been doing here and I was just wondering if you have any tips for me my summer test is next week and I'm struggling with
Trigonometry
Length area and volume
Geometry I