Author Topic: VCE Methods Question Thread! (Read 6048129 times) Tweet Share

Burt Macklin · « **Reply #11730 on:** August 11, 2015, 05:27:16 pm »

Hi, so I've been stuck on this part of a question for ages, would anyone be able to help?

A die is weighted as follows:
Pr(2)=Pr(3)=Pr(4)=Pr(5)=0.2
Pr(1)=Pr(6)=0.1

The die is rolled twice, and the smaller of the numbers showing is noted. Let Y represent this value

c. Find Pr(Y=1)

Floatzel98 · « **Reply #11731 on:** August 11, 2015, 06:00:46 pm »

Quote from: Burt Macklin on August 11, 2015, 05:27:16 pm

Hi, so I've been stuck on this part of a question for ages, would anyone be able to help?

A die is weighted as follows:
Pr(2)=Pr(3)=Pr(4)=Pr(5)=0.2
Pr(1)=Pr(6)=0.1

The die is rolled twice, and the smaller of the numbers showing is noted. Let Y represent this value

c. Find Pr(Y=1)

Well, you could just literally list the sample space for when Y = 1. If you look at each individual scenario you get 12 samples but 2 of them are the same since you can get (1,1) on both dies, so you minus one of those. Listing the whole sample space you get (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (1,1), (2,1), (3,1), (4,1), (5,1), (6,1). All the bold ones will have the same probabilities and the unbolded ones will have the same probabilities. So you eventually get $4(0.1*0.1) + 8(0.1*0.2) = 0.20$ . But remember you have to minus one of the double ups from the start, which gets you $0.20 - (0.1*0.1) = 0.19$

That's probably a hard, annoying way of looking at it. There is probably an easier way of thinking of it, but hopefully it helps

HopefulLawStudent · « **Reply #11732 on:** August 11, 2015, 06:07:06 pm »

Please help with part a)i)?

Burt Macklin · « **Reply #11733 on:** August 11, 2015, 06:12:02 pm »

Quote from: Floatzel98 on August 11, 2015, 06:00:46 pm

Well, you could just literally list the sample space for when Y = 1. If you look at each individual scenario you get 12 samples but 2 of them are the same since you can get (1,1) on both dies, so you minus one of those. Listing the whole sample space you get (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (1,1), (2,1), (3,1), (4,1), (5,1), (6,1). All the bold ones will have the same probabilities and the unbolded ones will have the same probabilities. So you eventually get $4(0.1*0.1) + 8(0.1*0.2) = 0.20$ . But remember you have to minus one of the double ups from the start, which gets you $0.20 - (0.1*0.1) = 0.19$

That's probably a hard, annoying way of looking at it. There is probably an easier way of thinking of it, but hopefully it helps

THANK YOU! It makes sense now.

Adiamond · « **Reply #11734 on:** August 11, 2015, 07:54:49 pm »

The other person was very helpful but that was such a tedious method!
Think of it like this: the lowest number shown can be '1' in three different ways: (1,1) , (1,0) , (0,1) with '0' being all of the possible numbers except for '1'. These probabilities can be calculated by multiplying the probability of the relative number on each role, e.g. a role of (1,1) would be [(0.1) * (0.1)] equaling 0.01, meaning that the probability of both rolls showing '1' is 0.01. Knowing this we can work out the probability of the first roll being '1' and the second roll being '0' (anything but '1'). The probability of all of the numbers that aren't 1 is 0.9, meaning that the probability of the roll being (1,0) is [(0.1) * (0.9)] and using symmetry the probability of rolling a (0,1) will be the same answer, 0.09.

Now that we have the probabilities of a '1' showing up in either of the two rolls, all that is left is to add the numbers up, being 0.01 + 0.09 + 0.09 = 0.19.

The other way works just as well but i thought that i would just share this way of doing it as well.

HopefulLawStudent · « **Reply #11735 on:** August 11, 2015, 08:21:19 pm »

Anyone?

Quote from: HopefulLawStudent on August 11, 2015, 06:07:06 pm

Please help with part a)i)?

Floatzel98 · « **Reply #11736 on:** August 12, 2015, 07:27:05 pm »

Given $Pr(A) = \frac{1}{2}$ , $Pr(B) = \frac{1}{4}$ and $Pr(A|B) = \frac{1}{6}$ , find:

c) $Pr(A'|B)$
d) $Pr(A|B')$

Bit confused on these 2, if anyone could help me out.

Zealous · « **Reply #11737 on:** August 12, 2015, 08:49:54 pm »

Quote from: Floatzel98 on August 12, 2015, 07:27:05 pm

Given $Pr(A) = \frac{1}{2}$ , $Pr(B) = \frac{1}{4}$ and $Pr(A|B) = \frac{1}{6}$ , find:

c) $Pr(A'|B)$
d) $Pr(A|B')$

Bit confused on these 2, if anyone could help me out.

Not a common method, but I think it's super easy to understand. You can create a karnaugh table using the given information:

$\left[\begin{array}{cccc} & A & A' & \\ B & ? & ? & \frac{1}{4} \\ B' & ? & ? & \frac{3}{4} \\ & \frac{1}{2} & \frac{1}{2} & 1 \end{array}\right]$

Then solve for Pr(A and B) using conditional probability:

$Pr(A|B)=\frac{Pr(A \ and \ B)}{Pr(B)} \implies \frac{1}{6}=\frac{Pr(A \ and \ B)}{\frac{1}{4}} \implies Pr(A \ and \ B)=\frac{1}{24}$

Then finish the table:

$\left[\begin{array}{cccc} & A & A' & \\ B & \frac{1}{24} & \frac{5}{24} & \frac{1}{4} \\ B' & \frac{11}{24} & \frac{7}{24} & \frac{3}{4} \\ & \frac{1}{2} & \frac{1}{2} & 1 \end{array}\right]$

Now you can read straight off the table and calculate your desired values!

Spoiler

$Pr(A'|B)=\frac{\frac{5}{24}}{\frac{1}{4}}=\frac{5}{6}$

$Pr(A|B')=\frac{\frac{11}{24}}{\frac{3}{4}}=\frac{11}{18}$

Edit: Woah I reached 700 posts I'm so old.

Floatzel98 · « **Reply #11738 on:** August 12, 2015, 08:59:09 pm »

Quote from: Zealous on August 12, 2015, 08:49:54 pm

Not a common method, but I think it's super easy to understand. You can create a karnaugh table using the given information:

$\left[\begin{array}{cccc} & A & A' & \\ B & ? & ? & \frac{1}{4} \\ B' & ? & ? & \frac{3}{4} \\ & \frac{1}{2} & \frac{1}{2} & 1 \end{array}\right]$

Then solve for Pr(A and B) using conditional probability:

$Pr(A|B)=\frac{Pr(A \ and \ B)}{Pr(B)} \implies \frac{1}{6}=\frac{Pr(A \ and \ B)}{\frac{1}{4}} \implies Pr(A \ and \ B)=\frac{1}{24}$

Then finish the table:

$\left[\begin{array}{cccc} & A & A' & \\ B & \frac{1}{24} & \frac{5}{24} & \frac{1}{4} \\ B' & \frac{11}{24} & \frac{7}{24} & \frac{3}{4} \\ & \frac{1}{2} & \frac{1}{2} & 1 \end{array}\right]$

Now you can read straight off the table and calculate your desired values!

Spoiler
$Pr(A'|B)=\frac{\frac{5}{24}}{\frac{1}{4}}=\frac{5}{6}$

$Pr(A|B')=\frac{\frac{11}{24}}{\frac{3}{4}}=\frac{11}{18}$

Edit: Woah I reached 700 posts I'm so old.

That makes everything so much easier. Thanks so much!!

Apink! · « **Reply #11739 on:** August 13, 2015, 11:16:46 am »

Hi,

I was wondering if anyone could help me with this question.

I got p=0, 4/7, 1

But the correct answer is 4/7 (only) Anyone mind telling me why 0 and 1 was not feasible?

Thank you

keltingmeith · « **Reply #11740 on:** August 13, 2015, 01:23:33 pm »

Quote from: Apink! on August 13, 2015, 11:16:46 am

Hi,

I was wondering if anyone could help me with this question.

I got p=0, 4/7, 1

But the correct answer is 4/7 (only) Anyone mind telling me why 0 and 1 was not feasible?

Thank you

The only logical reason I can find is because they're silly.

knightrider · « **Reply #11741 on:** August 13, 2015, 05:07:30 pm »

For this image attached.

How did they get rid of the 9 in step 2.

timton · « **Reply #11742 on:** August 13, 2015, 07:04:28 pm »

Hi everyone

Just wondering if there is a correct notation we should be using for Inverse Normal, as we cannot use invNorm as it is calculator notation. I know that normal distribution is N~(mean, var) but what is it for inverse?

Thanks everyone

cosine · « **Reply #11743 on:** August 13, 2015, 07:08:05 pm »

Quote from: knightrider on August 13, 2015, 05:07:30 pm

For this image attached.

How did they get rid of the 9 in step 2.

$9(x-\frac{2}{3})^2$
$= 3^2(x-\frac{2}{3})^2$
$= (3x^2(x-\frac{2}{3})^2)$
$= (3x-2)^2$

knightrider · « **Reply #11744 on:** August 13, 2015, 07:12:58 pm »

Quote from: cosine on August 13, 2015, 07:08:05 pm

$9(x-\frac{2}{3})^2$
$= 3^2(x-\frac{2}{3})^2$
$= (3x^2(x-\frac{2}{3})^2)$
$= (3x-2)^2$

Thankyou so much

but i cant see what you have done.

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