# From probability to proving

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• January 5th 2010, 03:37 AM
Mukilab
From probability to proving
Lots of questions that I've tried and gotten stuck on. I'll post the questions and my answers to them

1) Fully simplify $\frac{x^2-11x+28}{3x^2-10x-8}$
I got $\frac{(x-4)(x-7)}{(3x-4)(x+2)}$ I think that's right.

2) Nicolas has 30 beads in a bag. 9 of them are white. 8 green. 13 blue. What is the probability that if two beads are taken from the bag at random that they will not be the same colour. I got $\frac{113}{174}$ Definitely wrong

3)OPR is a triangle. $\longrightarrow$ OP = p and $\longrightarrow$OR = r
a.Find the vector $\longrightarrow$PR in terms of p and r
(No answer, I tried it but I just got stuck)

A os a point on PR such that PA:AR=7:2
b. Show that $\longrightarrow$OA = $\frac{1}{9}(2p+7r)$
Again I tried but couldn't even get close to an answer...

4)Prove that, for all positive integer values of n, $(7n+1)^2-(7n-1)^2$ is a multiple of 2.
Here I found out why but I just don't know how to word it so an exam will accept it...
• January 5th 2010, 03:59 AM
Archie Meade
$For\ Q1\ you\ have\ incorrectly factorised\ the\ denominator$

$-4\ and\ 2\ are\ interchanged.$

$For\ Q2,\ you\ should\ calculate\ 1-\frac{9c_2+8c_c+13c_2}{30c_2}$
• January 5th 2010, 04:09 AM
Archie Meade
$For\ Q3, draw\ your\ vector\ diagram\ with\ O\ at\ the\ origin.$

$Now\ pick\ 2\ points\ P\ and\ R.\ Draw\ the\ triangle.$

$Plot\ A\ near\ to\ R.$

$Vector\ pr=r-p\ and\ pa=\frac{7}{9}(r-p).$

$Then\ a-p=\frac{7}{9}(r-p)$

$a=\frac{7}{9}r-\frac{7}{9}p+\frac{9}{9}p$
• January 5th 2010, 04:17 AM
Archie Meade
$For\ Q4,\ you\ only\ need\ to\ show\ that\ the\ answer\ you\ get,$

$which\ is\ a\ multiple\ of\ n,\ can\ be\ written\ as\ (2n)(value)$
• January 5th 2010, 04:28 AM
Mukilab
Quote:

Originally Posted by Archie Meade
$For\ Q1\ you\ have\ incorrectly factorised\ the\ denominator$

$-4\ and\ 2\ are\ interchanged.$

$For\ Q2,\ you\ should\ calculate\ 1-\frac{9c_2+8c_c+13c_2}{30c_2}$

oh ok, I just realised ^^

I don't get how you got the second question

Quote:

Originally Posted by Archie Meade
$For\ Q4,\ you\ only\ need\ to\ show\ that\ the\ answer\ you\ get,$

$which\ is\ a\ multiple\ of\ n,\ can\ be\ written\ as\ (2n)(value)$

I also don't get what you mean here. Something like (2x2)(56)?
• January 5th 2010, 04:49 AM
Archie Meade
$For\ the\ 2nd\ one,\ you\ can\ calculate\ the\ probability\ of\ both\ being\ the\ same\ colour.$

$This\ is\ convenient,\ therefore\ there\ are\ 9c_2\ ways\ to\ pick\ 2\ white\ beads,$

$8c_2\ ways\ to\ pick\ 2\ green\ ones\ and\ 13c_2\ ways\ to\ pick\ 2\ blue\ ones.$

$If\ we\ add\ those\ up\ we've\ counted\ all\ the\ ways\ to\ pick\ 2\ beads\ of\ the\ same\ colour.$

$If\ we\ divide\ by\ the\ number\ of\ ways\ of\ picking\ any\ 2\ beads,$

$we\ have\ the\ probability\ of\ picking\ 2\ same\ colour\ beads.$

$All\ probabilities\ sum\ to\ 1, so\ if\ we\ subtract\ that\ probability$

$from\ 1, we\ have\ the\ probability\ of\ picking\ 2\ different\ colour\ beads.$

$In\ Q4,\ after\ simplifying\ your\ expression\ by\ multiplying\ out\ and\ subtracting,$

$your\ final\ answer\ is\ a\ multiple\ of\ n.$

$You\ now\ only\ need\ show\ that\ 2\ divides\ into\ this\ multiple.$

$If\ n\ is\ an\ integer,\ then\ 2n\ is\ a\ multiple\ of\ 2,\ 4n\ is\ a\ multiple\ of\ 2,$

$6n\ is\ a\ multiple\ of\ 2\ and\ so\ on.$
• January 5th 2010, 06:53 AM
Mukilab
Ok thank you for explaining the second but I am still confused to the first one. Why do you label certain probabilities as something such as 8c-subscript-2?

I'm quite surprised with your answer because this was quite an early question in a GCSE paper and probably the hardest one looking at the answer.
• January 5th 2010, 07:02 AM
Archie Meade
You might be more familiar with the notation $\binom{8}{2}$.

It means the number of ways to choose 2 from 8.

$\binom{8}{2}=\frac{8!}{(8-2)!2!}$

You must count the number of ways something can happen first before calculating the probability.
Here, we can count the opposite easily first.
• January 5th 2010, 07:28 AM
Mukilab
Errr, could someone else reiterate the whole answer? I tried it by drawing various lines and giving them each a probability...
• January 5th 2010, 07:37 AM
Archie Meade
You probably have $Nc_r$ and $Np_r$ buttons
on your calculator.

$Np_r$ will calculate the number of ways of arranging r items from a total of N.

$Nc_r$ will count the number of ways of selecting (without arranging) r items from N.

Using those keys is the fastest way to answer.
The probability of selecting 2 beads the same is $\frac{selections\ of\ 2\ same}{selections\ of\ 2\ from\ 30}$

The probability of the beads being different is that subtracted from 1.
• January 5th 2010, 07:42 AM
Mukilab
Quote:

Originally Posted by Archie Meade
You probably have $Nc_r$ and $Np_r$ buttons
on your calculator.

$Np_r$ will calculate the number of ways of arranging r items from a total of N.

$Nc_r$ will count the number of ways of selecting (without arranging) r items from N.

Using those keys is the fastest way to answer.
The probability of selecting 2 beads the same is $\frac{selections\ of\ 2\ same}{selections\ of\ 2\ from\ 30}$

The probability of the beads being different is that subtracted from 1.

Thank you for telling me those keys, very useful but I don't understand the core thing. I'll review this in a day or so and see if I still understand it
• January 5th 2010, 08:16 AM
Archie Meade
Someone might add something to the thread,
but once you know what those keys give you,
these calculations do become a snap.

Very adviseable to learn,
as those keys are there for that reason.
• January 5th 2010, 08:33 AM
Archie Meade
Here's an alternative...

Probability of 2 whites in a row is $\frac{9}{30}\ \frac{9-1}{30-1}=\frac{(9)(8)}{(30)(29)}$

Probability of 2 greens in a row is $\frac{8}{30}\ \frac{8-1}{30-1}=\frac{(8)(7)}{(30)(29)}$

Probability of 2 blues in a row is $\frac{13}{30}\ \frac{13-1}{30-1}=\frac{(13)(12)}{(30)(29)}$

The probability of getting 2 beads the same colour is the sum of these 3 probabilities, since we can get 2 whites, 2 greens or 2 blues.

Since all the probabilities of the situation of drawing 2 beads must sum to 1, all you need do is subtract the sum of those 3 from 1,
in order to find the probability that the first 2 beads will be of different colours.

Remember, if the first bead is white, there are 29 beads left and 8 are white,
ie one less bead and one less white.
Same for the others.
• January 5th 2010, 08:36 AM
Mukilab
Now I understand this, I used the same method apart from subtracting from one (so I obviously got it wrong).

Thank you for taking the time and consideration to think of another method that I will understand.
• January 5th 2010, 08:43 AM
Archie Meade
It might have helped if I'd checked how you got that value!

However, this was useful in that there is an alternative with those calculator buttons, especially when you need to check something like 17 out of 30.

For large numbers the keys are very very fast.

Thank you for your patience too, good work.