Maths SL P2 TZ2

[DeBrogliez]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Edited May 5, 2012 by DeBrogliez

[IBsurvivor2012]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

[Vincent Casey F]

this is strange.. I found paper 2 to be easier.. I used my GDC for almost everything!

matrices took me 2 minutes cause it was meant to be a GDC question

find a, b, and c was GDC too! You had to use PolySmlt2 for that - don't remember the answers but c=5

probability question was BinomialPDF for X=1 and X=2 (cause it said "at least 1 and at most 2")

acceleration was -31

the 2cos theta question ended up being 1.22 for theta

[DeBrogliez]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

[IBsurvivor2012]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

Oh crap! I didn't think about that... X_X :'( ... Totally forgot! Will I get at least some points for 'method' or 'recognized what to do'?

[DeBrogliez]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

Oh crap! I didn't think about that... X_X :'( ... Totally forgot! Will I get at least some points for 'method' or 'recognized what to do'?

I'm not sure about how they correct it

But it was 4 marks

So I wouldn't be worried

[Vincent Casey F]

they aren't even "events".. this was a binomial probability question.. a completely different section of the syllabus

[DeBrogliez]

they aren't even "events".. this was a binomial probability question.. a completely different section of the syllabus

Binomial is basically used to find the probability of repeated events!

[zeqqi]

how did you guys solve the matrix one? A*C*A^-1, isn't that the same as A*A^-1C, thus C=B. it was 5 marks, so it probably couldn't be that easy.

[Vincent Casey F]

for matrices it ended up being: C=(A^-1)(B)(A)

[zeqqi]

seems reasonable, in other words i failed lol. paper 2 wasn't what I expected.. it was pretty hard! even though i was able to answer most of the questions some of them were hard. anybody experience the same? i was scoring 80s on the past papers, and i literally hit the wall on this one.

Edited May 5, 2012 by zeqqi

[Ms. Procrastinator]

Does anyone know what it takes to actually get a 1 in Math SL? I've heard it's hard, like you have to do REALLY bad... still scared though.

[zeqqi]

can't be done.

[Ms. Procrastinator]

I hope so... don't know im just terrified about my paper 2! I was really ill and completely blanked, so i think i got max 20 points... Paper 1 was alright though got like 40 points and my math portfolio is a seven. Still im so sure my P2 is gonna make me fail

[Jyrgen]

Wasn't the matrix question like this:

A times C times inverse of A = B

How did you people solve it?

I removed the A from the front of C, while putting an inverse of A in front of B - can you do that?

Then I got rid of the inverse of A after C, by multiplying the right hand matrix by an A after it

Is that the answer?

That's what I did as well, it should be the right answer, I mean the method is legit so can't see how it could be wrong

Cheers

Let's just cross our fingers everyone and chill - no point in worrying, what's been done has been done

[uusinjsh]

I got 1/7, 5/21 and 1/5, if I recall correctly. First one being that 2 white balls are drawn from bag A, second one - that there will be 2 white balls from either bag A or bag B; third one - given there are 2 white balls, prob that they come from bag A

[uusinjsh]

As for binomials. My way of thinking was the following. At least one faulty... Only scenario in which this doesn't occur is when all are good. Prob of not being faulty = 0.95.

None not faulty? (0.95)^30. Then I subtracted the result from 1 and got 70-smth %.

Part B at least one faulty, not more than two is like X=1 and X=2. Binomial pdf for both, added and got the result.

Part C, the conditional. Answer from part B divided by answer from part A.

Edited May 5, 2012 by uusinjsh

[Jyrgen]

I got 1/7, 5/21 and 1/5, if I recall correctly. First one being that 2 white balls are drawn from bag A, second one - that there will be 2 white balls from either bag A or bag B; third one - given there are 2 white balls, prob that they come from bag A

I think I got the same

[Eastcoast93]

second binomial part was conditional probability btw. so P(AnB)/ P(B). paper was quite challenging imo as well. i messed up sequences and series question. i was always able to do these questions and then on the actual exam i blank on the question, maybe i get methods marks though for follow through (i made up an answer). made up the third equation for the matrix question on section B as well. its exactly like the population IA we did but i just couldn't think of the third equation, i knew perfectly well how to solve the equations. the final question on circular trigonometry confused me. didn't realise you had to use identities :/ couldn't find l either....but for the second part I graphed the theta graph and found 1.22 to be the root. shoudl picl up some method marks along the way. overall it was allright. harder than past years but it was decent i assume. paper 1 was really good so overall it should work out for me

[uusinjsh]

Oh, and the ratio of the sequence. I got 0.4 was that right?