# From probability to proving

• Jan 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 $\displaystyle \frac{x^2-11x+28}{3x^2-10x-8}$
I got $\displaystyle \frac{(x-4)(x-7)}{(3x-4)(x+2)}$ 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 $\displaystyle \frac{113}{174}$ Definitely wrong

3)OPR is a triangle. $\displaystyle \longrightarrow$ OP = p and $\displaystyle \longrightarrow$OR = r
a.Find the vector $\displaystyle \longrightarrow$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 $\displaystyle \longrightarrow$OA = $\displaystyle \frac{1}{9}(2p+7r)$
Again I tried but couldn't even get close to an answer...

4)Prove that, for all positive integer values of n, $\displaystyle (7n+1)^2-(7n-1)^2$ 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...
• Jan 5th 2010, 03:59 AM
$\displaystyle For\ Q1\ you\ have\ incorrectly factorised\ the\ denominator$

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

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

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

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

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

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

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

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

$\displaystyle For\ Q1\ you\ have\ incorrectly factorised\ the\ denominator$

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

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

oh ok, I just realised ^^

I don't get how you got the second question

Quote:

$\displaystyle For\ Q4,\ you\ only\ need\ to\ show\ that\ the\ answer\ you\ get,$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle 6n\ is\ a\ multiple\ of\ 2\ and\ so\ on.$
• Jan 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.
• Jan 5th 2010, 07:02 AM
You might be more familiar with the notation $\displaystyle \binom{8}{2}$.

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

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

You must count the number of ways something can happen first before calculating the probability.
Here, we can count the opposite easily first.
• Jan 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...
• Jan 5th 2010, 07:37 AM
You probably have $\displaystyle Nc_r$ and $\displaystyle Np_r$ buttons

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

$\displaystyle Nc_r$ 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 $\displaystyle \frac{selections\ of\ 2\ same}{selections\ of\ 2\ from\ 30}$

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

You probably have $\displaystyle Nc_r$ and $\displaystyle Np_r$ buttons

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

$\displaystyle Nc_r$ 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 $\displaystyle \frac{selections\ of\ 2\ same}{selections\ of\ 2\ from\ 30}$

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
• Jan 5th 2010, 08:16 AM
but once you know what those keys give you,
these calculations do become a snap.

as those keys are there for that reason.
• Jan 5th 2010, 08:33 AM
Here's an alternative...

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

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

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

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.
• Jan 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.
• Jan 5th 2010, 08:43 AM