Author Topic: VCE Methods Question Thread! (Read 5689581 times) Tweet Share

Zealous · « **Reply #2310 on:** July 30, 2013, 07:59:44 pm »

Hey guys, just got a few probability questions =)
All help is greatly appreciated! tyvm

Question 1: Independent Events

Spoiler

In a random experiment, a fair six sided die is rolled and the upward facing number recorded. Two
independent events could be:
A. {1,2,3} and {4,5,6}
B. {1,3} and {2,3,6}
C. {1,5} and {1,3}
D. {3,5} and {1,3,5}
E. {1} and {2}

Okay, I've seen these questions around a lot and not been sure on how to approach them.
Independent events state that $Pr(X \cap Y)=Pr{(X)}\times{Pr(Y)}$

How am I supposed to figure out what the intersection is with these sort of events? or the individual probabilities if it's in a set like this?

Question 2: woo deck of cards

Spoiler

5 cards are drawn at random from a 52 card deck containing equal numbers of each suit (spades, clubs,
hearts, and diamonds) without replacement. What is the probability that all the cards drawn are of the same suit?

The answer is $\frac{4 \times 13! \times 37!}{8! \times 52!}$

Could someone just quickly run me through how they've come to this?

Question 3: Charles and Callum

Spoiler

Callum and Charles meet every week for a game of ping pong. Disappointed with his recent losing
streak, Callum tries to identify the cause of his problems. Being an emotional player, Callum feels that the
result of his previous game affects his performance. He decides that his probability of winning given that
he won the last game is 0.6, while his probability of winning given that he lost the last game is only 0.3.

Find the exact proportion of games Callum is expected to win in the long run if the probability he wins the first match is 0.5.

So we can create a transition matrix: (where W1 is win on last game)

$M = \bordermatrix{~ & W1 & L1 \cr W2 & 0.6 & 0.3 \cr L2 & 0.4 & 0.7 \cr}$

From here, I raise the transition matrix to a very high power and multiplied it by a 2 x 1 matrix with 0.5 and 0.5.
I got the answer which is 0.428571. Although the answer wanted it in exact, solutions showed that all they did was:
$\frac{0.3}{0.4+0.3}=\frac{3}{7}$ .. haven't figure out why yet though.. any ideas where they got those figures?

Thank you very much! haha some of the probability stuff messes with my head, and thanks for reading such a long post (if you did :0)

saba.ay · « **Reply #2311 on:** July 30, 2013, 08:16:55 pm »

This might be a very broad/vague/difficult to explain question, but could someone please explain Combinations and how it/they work? Alternatively, if anybody knows of good links which explain combinations well, I'd love to check them out.

Please and thanks

Alwin · « **Reply #2312 on:** July 30, 2013, 08:48:57 pm »

Quote from: sushi. on July 30, 2013, 07:59:44 pm

Hey guys, just got a few probability questions =)
All help is greatly appreciated! tyvm

Question 1: Independent Events
Spoiler
In a random experiment, a fair six sided die is rolled and the upward facing number recorded. Two
independent events could be:
A. {1,2,3} and {4,5,6}
B. {1,3} and {2,3,6}
C. {1,5} and {1,3}
D. {3,5} and {1,3,5}
E. {1} and {2}

Okay, I've seen these questions around a lot and not been sure on how to approach them.
Independent events state that $Pr(X \cap Y)=Pr{(X)}\times{Pr(Y)}$

How am I supposed to figure out what the intersection is with these sort of events? or the individual probabilities if it's in a set like this?
Question 2: woo deck of cards
Spoiler
5 cards are drawn at random from a 52 card deck containing equal numbers of each suit (spades, clubs,
hearts, and diamonds) without replacement. What is the probability that all the cards drawn are of the same suit?

The answer is $\frac{4 \times 13! \times 37!}{8! \times 52!}$

Could someone just quickly run me through how they've come to this?

Question 3: Charles and Callum
Spoiler
Callum and Charles meet every week for a game of ping pong. Disappointed with his recent losing
streak, Callum tries to identify the cause of his problems. Being an emotional player, Callum feels that the
result of his previous game affects his performance. He decides that his probability of winning given that
he won the last game is 0.6, while his probability of winning given that he lost the last game is only 0.3.

Find the exact proportion of games Callum is expected to win in the long run if the probability he wins the first match is 0.5.

So we can create a transition matrix: (where W1 is win on last game)

$M = \bordermatrix{~ & W1 & L1 \cr W2 & 0.6 & 0.3 \cr L2 & 0.4 & 0.7 \cr}$

From here, I raise the transition matrix to a very high power and multiplied it by a 2 x 1 matrix with 0.5 and 0.5.
I got the answer which is 0.428571. Although the answer wanted it in exact, solutions showed that all they did was:
$\frac{0.3}{0.4+0.3}=\frac{3}{7}$ .. haven't figure out why yet though.. any ideas where they got those figures?
Thank you very much! haha some of the probability stuff messes with my head, and thanks for reading such a long post (if you did :0)

Q1

What we do is denote the sets. I'll run though option A with you,
let P(X) be probability of getting something in the set {1, 2, 3}. So P(X) = 1/2
let P(Y) be probability of getting something in the set {4, 5, 6}. So P(Y) = 1/2
This implies that P(X n Y) is the probability of getting something in the set {{1, 2, 3} n {4, 5, 6}} = null set. So P(X n Y) = 0
Use the rule P(X n Y) = P(X) x P(Y) for independent events:
LHS = 0, but RHS = 1/2 * 1/2 = 1/4. LHS ≠ RHS, this option is not an independent event.

Answer

C, since LHS = 1/4 and RHS = 1/2 * 1/2

Q2 EDIT: thanks nliu! didn't read the q, i worked out choosing 4.. not 5 cards..

Okay, so lets SUPPOSE the suit is hearts <3. Find the probability of choosing 5 from the hearts. So:
$\text{Probability} = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \times \frac{10}{49} \times \frac{9}{48}= \frac{9 \times 10 \times 11 \times 12 \times 13}{48 \times 49 \times 50 \times 51 \times 52}$
Now here's the a trick:
$\frac{9 \times 10 \times 11 \times 12 \times 13}{48\times 49 \times 50 \times 51 \times 52} = \frac{\frac{13!}{8!}}{\frac{52!}{47!}}= \frac{13! \times 47!}{52! \times 8!}$

Now, what if we had chosen one of the other 4 suits, not just hearts. So then, its:

Pr(choose 4 hearts) + Pr(choose 4 spades) + Pr(choose 4 diamonds) + Pr(choose 4 clubs)
$\frac{13! \times 47!}{52! \times 8!} + \frac{13! \times 47!}{52! \times 8!} + \frac{13! \times 47!}{52! \times 8!} + \frac{13! \times 47!}{52! \times 8!} = 4\times \frac{13! \times 47!}{52! \times 8!}$

Q3

It uses a steady state short cut.
$T=\begin{pmatrix} 1-b & a \\ b & 1-a \end{pmatrix}$
Steady state for Pr(a) is:
$\frac{a}{a+b}$

EDIT

I can guess your question from Q3, why does it work?
S_n+1 = T S_n, even when you times by T again, it makes no difference in the STEADY state. So:
$\begin{alignedat}{1} \begin{pmatrix} A \\ B \end{pmatrix} &= \begin{pmatrix} 1-b & a \\ b & 1-a \end{pmatrix} \times \begin{pmatrix} A \\ B \end{pmatrix} \\ \text{But also, } A+B &= 1 \end{alignedat}$

Solving give the "shortcut formula"

Hope it helps!

Professor Polonsky · « **Reply #2313 on:** July 30, 2013, 08:51:57 pm »

Code: [Select]

\heartsuit \diamondsuit \clubsuit \spadesuit
$\heartsuit \diamondsuit \clubsuit \spadesuit$

LaTeX magic

Zealous · « **Reply #2314 on:** July 30, 2013, 08:59:19 pm »

Quote from: Alwin on July 30, 2013, 08:48:57 pm

Q1
What we do is denote the sets. I'll run though option A with you,
let P(X) be probability of getting something in the set {1, 2, 3}. So P(X) = 1/2
let P(Y) be probability of getting something in the set {4, 5, 6}. So P(Y) = 1/2
.......... [Edited out some of Alwin's working]
Q3
It uses a steady state short cut.
$\begin{pmatrix} 1-b & a \\ b & 1-a \end{pmatrix}$
Steady state for Pr(a) is:
$\frac{a}{a+b}$

Hope it helps!

Thanks Alwin (I was about to call you Alvin again)! Very helpful, I understand em now, quite smart little tricks.

Are there any other steady state shortcuts? Is there one to find the steady state of b?
Also, for steady states "in the long run", it doesn't matter what the initial state is right? as eventually the probabilities will even out?

lzxnl · « **Reply #2315 on:** July 30, 2013, 09:00:05 pm »

48!/9! is not 37!. Just saying. Think about it. 48!/9! is still divisible by 41. 37! isn't.
Alwin, you've answered a question for "four cards of the same suit". Question says 5.

So...here we go.
How many ways can you pick 5 cards from a particular suit? 13*12*11*10*9. or 13!/8!. This isn't taking into account repetitions of these 5 cards, so 13!/8!5! (you can arrange n individual objects in n! ways)

How many ways can you pick 5 cards from a deck of 52? Using a similar logic, 52!/47!5!

So divide the first by the second and multiply by four to account for all suits.

13!*47!*4/8!52!

I do believe you've made a typo there in what you've typed as an answer.

Alwin · « **Reply #2316 on:** July 30, 2013, 09:07:09 pm »

Quote from: nliu1995 on July 30, 2013, 09:00:05 pm

48!/9! is not 37!. Just saying. Think about it. 48!/9! is still divisible by 41. 37! isn't.
Alwin, you've answered a question for "four cards of the same suit". Question says 5.

So...here we go.
How many ways can you pick 5 cards from a particular suit? 13*12*11*10*9. or 13!/8!. This isn't taking into account repetitions of these 5 cards, so 13!/8!5! (you can arrange n individual objects in n! ways)

How many ways can you pick 5 cards from a deck of 52? Using a similar logic, 52!/47!5!

So divide the first by the second and multiply by four to account for all suits.

13!*47!*4/8!52!

I do believe you've made a typo there in what you've typed as an answer.

You're correct nliu, I realised that soon after I posted (editing now), but my laptop decided it would be a perfect time to run out of battery -.- THANKS!

Alwin · « **Reply #2317 on:** July 30, 2013, 09:22:41 pm »

Quote from: sushi. on July 30, 2013, 08:59:19 pm

Thanks Alwin (I was about to call you Alvin again)! Very helpful, I understand em now, quite smart little tricks.

Are there any other steady state shortcuts? Is there one to find the steady state of b?
Also, for steady states "in the long run", it doesn't matter what the initial state is right? as eventually the probabilities will even out?

Yup, initial don't matter

Quote from: sabzie. on July 30, 2013, 08:16:55 pm

This might be a very broad/vague/difficult to explain question, but could someone please explain Combinations and how it/they work? Alternatively, if anybody knows of good links which explain combinations well, I'd love to check them out.

Please and thanks

Haha, you could be right about the broadness/vagueness/difficultiness of this q, but I can try give it a shot. I assume you have looked at whatever proofs are included in your textbook already ??
$^n C _r = \frac{^n P _r}{r!} = \frac{n!}{(n-r)! r!}$

So why do we divide by r! ? we do this because we are only interested in combinations, not the order (permutation).
If we were choosing 5 people from a group of 20, the ordered choice would be ²⁰P₅
BUT, the 5 people can be chosen in 5! ways, which gives the order (permutation). If we want to get rid of the order (find the combination), we divide by 5!.
Hence, the rule above

Hope it helps!!

chemdeath · « **Reply #2318 on:** July 31, 2013, 07:57:02 pm »

Need help with this question ! Thanks

Find average speed over given interval for following speed fn
Time is in secs and speed in meters per sec

V=24sin(1/4pi*t) over interval [0,4]

Phy124 · « **Reply #2319 on:** July 31, 2013, 08:09:43 pm »

Quote from: chemdeath on July 31, 2013, 07:57:02 pm

Need help with this question ! Thanks

Find average speed over given interval for following speed fn
Time is in secs and speed in meters per sec

V=24sin(1/4pi*t) over interval [0,4]

The average value of a function, $f(x)$ , over the interval $[a,b]$ , is given by:

$\bar{f} = \frac{1}{b-a}\int^b_a f(x) \ dx$

So the average speed in this case is given by:

$\bar{v} = \frac{1}{4-0}\int^4_0 24\sin\left(\frac{\pi t}{4}\right)dt = \frac{48}{\pi}$ m/s

saba.ay · « **Reply #2320 on:** August 01, 2013, 08:01:10 am »

Quote from: Alwin on July 30, 2013, 09:22:41 pm

So why do we divide by r! ? we do this because we are only interested in combinations, not the order (permutation).
If we were choosing 5 people from a group of 20, the ordered choice would be ²⁰P₅
BUT, the 5 people can be chosen in 5! ways, which gives the order (permutation). If we want to get rid of the order (find the combination), we divide by 5!.
Hence, the rule above

Hope it helps!!

Thank you. That helped.

darklight · « **Reply #2321 on:** August 01, 2013, 05:08:03 pm »

Can someone help me with part b) of this question? Thanks!

Homer · « **Reply #2322 on:** August 01, 2013, 05:48:02 pm »

Quote from: darklight on August 01, 2013, 05:08:03 pm

Can someone help me with part b) of this question? Thanks!

heyy! just a total guess but is the x co ordinate for b $\frac{-1+m}{n}$

darklight · « **Reply #2323 on:** August 01, 2013, 05:49:29 pm »

Quote from: Homer on August 01, 2013, 05:48:02 pm

heyy! just a total guess but is the x co ordinate for b $\frac{-1+m}{n}$

It is

How did you get there?

Jaswinder · « **Reply #2324 on:** August 01, 2013, 06:09:12 pm »

hey guys whats a good way of finding the mean of the distribution. ans: 0.5
thanks

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