Author Topic: VCE Methods Question Thread! (Read 5782188 times) Tweet Share

Phy124 · « **Reply #2520 on:** September 09, 2013, 06:14:59 pm »

Quote from: lolipopper on September 09, 2013, 04:59:20 pm

i meant that i wasn't sure whether i was wrong or right. im bad at methods.

Don't bother asking TT for help regarding a lack of confidence in ones answer, he is not familiar with this concept

jono88 · « **Reply #2521 on:** September 09, 2013, 06:16:39 pm »

The chance of winning a major prize in a raffle is 0.09. The number of tickets required to ensure a probability of more than 0.75 of winning a major prize at least once is:
A   14
B   15
C   16
D   17
E   18

RKTR · « **Reply #2522 on:** September 09, 2013, 06:35:42 pm »

Quote from: jono88 on September 09, 2013, 06:16:39 pm

The chance of winning a major prize in a raffle is 0.09. The number of tickets required to ensure a probability of more than 0.75 of winning a major prize at least once is:
A   14
B   15
C   16
D   17
E   18

probability of winning at least once needs to be more than 0.75
therefore probability of not winning needs to be less than 0.25

nC0 (0.09)^0 (0.91)^n < 0.25
0.91^n < 0.25
log (0.91)^n < log(0.25)
n log(0.91) < log(0.25)
n > log(0.25) / log(0.91) the < is changed to > because log(0.91) is negative
n > 14.6992
n=15 (B)

Alwin · « **Reply #2523 on:** September 09, 2013, 06:48:34 pm »

Quote from: jono88 on September 09, 2013, 06:16:39 pm

The chance of winning a major prize in a raffle is 0.09. The number of tickets required to ensure a probability of more than 0.75 of winning a major prize at least once is:
A   14
B   15
C   16
D   17
E   18

OKAY I BLOODY BAGS THIS ONE BECAUSE EVERYTIME SOMEONE WILL SAY, NO YOU CAN'T SOLVE THAT ON CAS.
BUT YOU CAN!!!
EDIT: Beaten

-end rant. begin solution-

$\begin{alignedat}{1} \text{let the number of trials be } n \text{ and } p &=0.09 \\ \text{let } X & \text{be winning a major prize} \\ \text{we require: } Pr(X \geq 1) &=0.75 \\ \text{equivalently, } Pr(X < 1) &= 0.25 \\ Pr(X=0) &= 0.25 \\ \begin{pmatrix} n \\ 0 \end{pmatrix} \times (0.91)^{n} (0.09)^{0} &= 0.25 \\ \text{ } \\ \text{now, we go back to the definition of } \begin{pmatrix} n \\ r \end{pmatrix} &= \frac{n!}{r!(n-r)!} \\ \therefore \begin{pmatrix} n \\ 0 \end{pmatrix} &= \frac{n!}{0!(n-0)!} \\ &= \frac{n!}{n!} = 1 \\ \text{ } \\ \begin{pmatrix} n \\ 0 \end{pmatrix} \times (0.91)^{n} (0.09)^{0} &= 0.25 \\ 1 \times (0.91)^{n} (0.09)^{0} &= 0.25 \\ \text{now, we can use cas to solve this, it's good at polynomials } &+ \text{ powers} \\ x &\approx 14.699 \\ \text{as always we round up, so } x &= 15 \end{alignedat}$

So, the answer is [ B ]

Now, we can apply this to harder questions that some people would argue: you have to make a table of values, or trial and error is needed, or just graph it You just need to break down the combinations

For example, it the had been winning a major prize at least three times:
$\begin{alignedat}{1} \text{let the number of trials be } Pr(X \geq 3) &=0.75 \\ \text{equivalently, } Pr(X < 3) &= 0.25 \\ Pr(X=0) + Pr(X=1) + Pr(X=2) &= 0.25 \\ \begin{pmatrix} n \\ 0 \end{pmatrix} \times (0.91)^{n} (0.09)^{0} + \begin{pmatrix} n \\ 1 \end{pmatrix} \times (0.91)^{n-1} (0.09)^{1} + \begin{pmatrix} n \\ 2 \end{pmatrix} \times (0.91)^{n-2} (0.09)^{2} &= 0.25 \\ \frac{n!}{0!(n-0)!} \times (0.91)^{n} (0.09)^{0} + \frac{n!}{1!(n-1)!} \times (0.91)^{n-1} (0.09)^{1} + \frac{n!}{2!(n-2)!} \times (0.91)^{n-2} (0.09)^{2} &= 0.25 \\ 1 \times (0.91)^{n} (0.09)^{0} + n \times (0.91)^{n-1} (0.09)^{1} + \frac{1}{2}n(n-1) \times (0.91)^{n-2} (0.09)^{2} &= 0.25 \\ \text{now, we can use cas to solve this, it's good at polynomials } &+ \text{ powers} \\ x &\approx 42.58 \\ \text{as always we round up, so } x &= 43 \end{alignedat}$

Hope it helps

jono88 · « **Reply #2524 on:** September 09, 2013, 09:27:20 pm »

Wow thanks guys!
Good use of an extra adaptation to the question there Alwin. xD
some people would argue: "you have to make a table of values, or trial and error is needed, or just graph it" LOL

jimmy22 · « **Reply #2525 on:** September 10, 2013, 08:57:44 pm »

At each of a series of trials, the probability of the occurrence of a certain event is 1/2, except that it cannot occur in two consecutive trials.

a) Show that the probability of it occurring just twice in three trials is 1/4.

I've tried using a tree diagram, and i keep getting 1/8. Could someone please show me how to do this.
Thanks

lzxnl · « **Reply #2526 on:** September 10, 2013, 09:03:04 pm »

Quote from: JM14 on September 10, 2013, 08:57:44 pm

At each of a series of trials, the probability of the occurrence of a certain event is 1/2, except that it cannot occur in two consecutive trials.

a) Show that the probability of it occurring just twice in three trials is 1/4.

I've tried using a tree diagram, and i keep getting 1/8. Could someone please show me how to do this.
Thanks

So...these are the probabilities. Let F be fail and S be success.
In total, you can have SFS, SFF, FSF, FFS, FFF as possibilities.
So...we have SFS only. Which makes sense as we can't have two in a row.
The probability of the first success is 1/2. The probability of the failure is 1 because that's all that is allowed after a success (two consecutive successes not allowed), while the probability of the last success is 1/2. Multiply to give 1/4.

jimmy22 · « **Reply #2527 on:** September 10, 2013, 09:11:16 pm »

oh okay, that makes a lot of sense about the pr(of a failure being 1) right after a success.

Thanks very much!

Zealous · « **Reply #2528 on:** September 13, 2013, 04:24:29 pm »

If it is raining on a particular day the chance that it will rain the next day is 0.7. If it is not raining on a particular day, the chance that it will rain on the following day is 0.4. Calculate the probability that, if it is raining on Monday, it will also be raining on Thursday of the same week.
Could someone just quickly run through that question if you have time. I think the answers are incorrect =p and just like someone else to confirm. Thanks!

BasicAcid · « **Reply #2529 on:** September 13, 2013, 04:32:34 pm »

Quote from: zealous on September 13, 2013, 04:24:29 pm

If it is raining on a particular day the chance that it will rain the next day is 0.7. If it is not raining on a particular day, the chance that it will rain on the following day is 0.4. Calculate the probability that, if it is raining on Monday, it will also be raining on Thursday of the same week.
Could someone just quickly run through that question if you have time. I think the answers are incorrect =p and just like someone else to confirm. Thanks!

[0.7 0.4]^3 [1] equals [0.583]
[0.3 0.6]

[0.417].

b^3 · « **Reply #2530 on:** September 13, 2013, 04:37:10 pm »

Let $B$ be the event that it rains today.
Let $A$ be the event that it rained on the previous day.
Then we have $\text{Pr}(B|A)=0.7$ and $\text{Pr}(B|A')=0.4$
Now we need our transition matrix,
$\begin{alignedat}{1}T & =\begin{bmatrix}0.7 & - \\ - & 0.6 \end{bmatrix}\end{alignedat}$
Now we also know that the columns of a transition matrix have to sum to 1, so we have
$\begin{alignedat}{1}T & =\begin{bmatrix}0.7 & 0.4 \\ 0.3 & 0.6 \end{bmatrix}\end{alignedat}$
We are told that it is raining on Monday, so we have our initial state, $\begin{alignedat}{1}S_{0} & =\begin{bmatrix}1 \\ 0 \end{bmatrix}\end{alignedat}$
So here Monday is $S_{0}$ which makes Thursday {tex]S_{3}[/tex].
$S_{n}=T^{n} \times S_{0}$
So, for Thursday
$\begin{alignedat}{1}S_{3} & =\begin{bmatrix}0.7 & 0.4 \\ 0.3 & 0.6 \end{bmatrix}^{3}\times\begin{bmatrix}1 \\ 0 \end{bmatrix} \\ & =\begin{bmatrix}0.5830 \\ 0.4170 \end{bmatrix} \end{alignedat}$
The top entry corresponds to a success, that is that it rains on Thursday, while the bottom entry corresponds to a failure, that is that it doesn't rain on Thursday, so we want the top entry.

The probability that it rains on Thursday is 0.5830.

EDIT: Beaten.

BasicAcid · « **Reply #2531 on:** September 13, 2013, 04:45:37 pm »

Quote from: b^3 on September 13, 2013, 04:37:10 pm

EDIT: Beaten.

Nah you actually explained it, I really like the way you actually explain each and every step which I've noticed you do in all of your posts in these maths sections.

So no, you beat me.

Zealous · « **Reply #2532 on:** September 13, 2013, 05:23:08 pm »

Quote from: b^3 on September 13, 2013, 04:37:10 pm

The probability that it rains on Thursday is 0.5830.

Quote from: BasicAcid on September 13, 2013, 04:32:34 pm

[0.7 0.4]^3 [1] equals [0.583]
[0.3 0.6]
[0.417].

Thanks guys, the solutions to the exam were wrong. I wasn't right though, I accidentally did 7^3 as 243.. =o aiyoh (Exam 1 btw).

It bug me when their answers are wrong, if I see no reason why their answer should be correct it's great that I can get a real quick confirmation of my answer over here.

Nato · « **Reply #2533 on:** September 13, 2013, 08:29:37 pm »

trig question here (1/2 btw)

how should i go about sketiching trig function without x-intercepts (such as $3\sin(2x)+2=0$

i mean when they do have x-intercepts, i find them, find the endpoints, y-int etc and sketch them.

Stevensmay · « **Reply #2534 on:** September 14, 2013, 12:45:12 am »

Quote from: Student0001 on September 13, 2013, 08:29:37 pm

how should i go about sketiching trig function without x-intercepts (such as $3\sin(2x)+2=0$

So seeing as it's not going to be that easy to find points that we can then sketch around, we will apply transformations over a series of graphs until we have our final result.

We need to apply our transformations in the order of DR T (easy to remember), so dilations and reflections then translations.
Firstly we start by graphing $\sin(2x)$
We can find the period of this with $Period = \frac{2 \pi}{b}=\frac{2 \pi}{2}$ , thus the period is $\pi$ . So now we would sketch this, which is a simple sin graph with a period of pi that we can easily do.

Next we dilate our sin(2x) graph by a factor of 3, so essentially multiply every y value by 3, and then regraph this. By now we should have a sin(2x) graph with an amplitude of 3.

Finally we apply the translation, which is a translation of +2 units in the y direction. Simply add 2 to all the y values, producing the correct translation.

We now have a 3sin(2x)+2 graph.

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