BEC'S methods questions

[bec]

1. semi-flip-floppy?
2. thanks, i sort of get it but Pr(bec remembers formulae)=0 lol. but it is good to understand it since in most of this chapter i'm just shoving things into formulae they give me in the summary
3. we get a formula sheet???

[Collin Li]

It's semi-flip-floppy because it's not a formal proof at all , it just provides the "gist of it."

You get a common formula sheet provided by VCAA.

2006 Exam 1: http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/pastexams/2006mmcas1-w.pdf
Look at pages 11-13.

You will get this for Exam 2 as well (not that you'll need it because you get to have an entire bound book reference for this exam - I hope you know that!)

[bec]

that's awesome that we get a formula sheet! i never knew! kind of weird that they have the area of a triangle which we all learnt in year 7, but not any trig identities...but still, it's pretty good

this question should be really easy but i can't do it:

A spinner has four equally likely sectors numbered from 1 to 4
Find Pr(X=3|X "is not equal to" 1)

The probability function is obviously
x       1    2   3   4
p(x) .25 .25 .25 .25

and i think the thing i'm most stuck on is calculating

help?

edit: changed the union to an intercept

[enwiabe]

YES FINALLY ONE THAT COBLIN/AHMAD/HUMPHDOGG HASN'T GOT TO FIRST! :d:d:d

N.B. "not equal to" may be represented by "!="

FIRSTLY:

Pr(A|B) = Pr(A INTERSECT B)/Pr(B) it's not the UNION of, as you put it.

Therefore,
Pr(X=3|X != 1) = Pr(X=3 intersect X != 1) / Pr(X != 1)

Now the intersection of Pr(X=3) and Pr(X != 1) is clearly Pr(X=3) as Pr(X=3) fits exactly into set one, and partially into set 2, being its intersection.

Pr(X = 3) = .25

Now, Pr(X != 1) = Pr(X=2) + Pr(X=3) + Pr(X=4) = .25 + .25 + .25 = .75

Therefore,
Pr(X=3|X != 1) = Pr(X=3)/Pr(X != 1) = .25/.75 = 1/3

Hope this was of help!

[bec]

gahhh those things always look so easy when someone else does them! thank you!

and yet another question....
If an experiment is performed once, the distribution of X has the probability function shown here.

x       0     1     2     3     4
p(x) .23   .27  .19  .13   .18

If the same experiment is performed several times, the probability of each of the outcomes would remain constant for each experiment.

If the experiment is performed five times, what is the probability that X>2 three times?

i realllly appreciate all this help i'm getting from everyone, thank you so much

[enwiabe]

Pr(X > 2) = Pr(X=3) + Pr(X=4) = .31

Now you use the binomial distribution, where p = .31, n = 5 and x = 3.

If you let R be the number of times that X > 2,
Pr(R = 3) = 5C3(.31)^3(.69)^2 = 10 x .31^3 x .69^2 = .142

[bec]

thanks

one more! i have no idea where to even start for this one.

A loaded die is tossed and the number X showing has the probability function shown in the table:

x              1                2          3         4          5            6
p(x)    (1-k)/10   1/10    1/5    3/10    3/10    k/10

a) Find the value of k for which the probability of obtaining a total of 7 in two tosses of this die is a maximum

I'll put up part b later if i still don't get it

[enwiabe]

Firstly, let us consider "the probability of obtaining a total of 7".

Let T = total obtained by two die rolls.

Considering Pr(T=7), that means that you need 1+6, 2+5 or 3+4. Conversely, this may be obtained with 6+1, 5+2, and 4+3.

This means that you have 6 possible outcomes in which you may get Pr(T=7).

Pr(T=7) = 2xPr(1&6) + 2xPr(3&4) + 2xPr(5&2)

Consider 1&6. Pr(1) = (1-k)/10 and Pr(6) = k/10

The probability of rolling a 1 and THEN a 6 is Pr(1) x Pr(6) = k(1-k)/10

therefore, Pr(1&6) = k(1-k)/100

Pr(2&5) = 3/10 x 1/10 = 3/100
Pr(3&4) = 1/5 x 3/10 = 3/50

Remember, Pr(T=7) = 2xPr(1&6) + 2xPr(2&5) + 2xPr(3&4)
                           = k(1-k)/50 + 3/50 + 3/25
                           = k(1-k)/50 + 9/50 
                           = (k - k^2)/50 + 9/50

We now have a formula for Pr(T=7)! But the question stipulates that it wants a MAXIMUM (here is where the question turns from a probability question into a calculus max/min problem)

Consider the graph of Pr(T=7) = y = (k-k^2)/50 + 9/50
                                              = (9 + k - k^2)/50

It is an inverted parabola. Thus, the maximum occurs when dy/dk = 0

So dy/dk = (1-2k)/50

Set dy/dk = 0

And you obtain 1 - 2k = 0 and k = 1/2

 

[bec]

thanks for that, it makes a lot of sense

the second part to this question is Let Y be the sum of the numbers obtained in two tosses of this die. For the value of k found in part a, find:
i) The mode of Y
ii) The median of Y
iii) The mean of Y

I think i can do this 'manually' by writing out the probability of each of the sums 2-12 being obtained:
eg.
Y=4, obtained by rolling 2,2; 3,1 or 1,3
Pr(Y=4) = Pr(X=2) x Pr(X=2) + Pr(X=3) x Pr(X=1) + Pr(X=1) x Pr(X=3)
           = 1/10 x 1/10 + 1/5 x 1/20 + 1/20 x 1/5
           = 0.03

But doing that for each of the sums/Y-values would be time-consuming. Is there a more efficient way of solving this?

[Ahmad]

This was brought up:

1.

Proof:

2.

Proof:

3.

Proof:

Hope it helps.

[Collin Li]

I love that square block!

It would replace my full stop if I could... actually, why the hell not?

[humph]

for proofs, it really should be a square block outline, not a filled square block 

/nitpicky.

[Ahmad]

I don't think there's any fixed convention there, everywhere I've seen it it's been a filled in square block!

Edit: See electronic forms here http://en.wikipedia.org/wiki/Q.E.D.

[humph]

I don't think there's any fixed convention there, everywhere I've seen it it's been a filled in square block!

Edit: See electronic forms here http://en.wikipedia.org/wiki/Q.E.D.

fair enough, i've only ever seen the hollow square or very rarely Q.E.D. guess it varies...

[/0]

thanks for that, it makes a lot of sense

the second part to this question is Let Y be the sum of the numbers obtained in two tosses of this die. For the value of k found in part a, find:
i) The mode of Y
ii) The median of Y
iii) The mean of Y

I think i can do this 'manually' by writing out the probability of each of the sums 2-12 being obtained:
eg.
Y=4, obtained by rolling 2,2; 3,1 or 1,3
Pr(Y=4) = Pr(X=2) x Pr(X=2) + Pr(X=3) x Pr(X=1) + Pr(X=1) x Pr(X=3)
           = 1/10 x 1/10 + 1/5 x 1/20 + 1/20 x 1/5
           = 0.03

But doing that for each of the sums/Y-values would be time-consuming. Is there a more efficient way of solving this?

Hi, I'd be interested to see how you can do these questions. I got 7 for the first two, but for the third I'm not being efficient enough.