Extension 1 predictions? (2 Viewers)

malcolm21 · Oct 26, 2015

iforgotmyname said:
Jim takes a dump every morning.

Probability of him taking a large one = 7/10, probability of him taking a small one =3/10. Calculate how many consecutive large dumps he would need to take so that he ensures his next dump is a small one

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Thanks, jim

iforgotmyname · Oct 26, 2015

Glad you like my question....

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Crisium · Oct 26, 2015

photastic said:
Hmm, they haven't added projectile motion into the multiple choice section yet.

The only trial paper I've seen projectile in is James Ruse

photastic · Oct 26, 2015

Crisium said:
The only trial paper I've seen projectile in is James Ruse

and CSSA.

malcolm21 · Oct 26, 2015

Ekman said:
My predictions:

Circle Geo: Either question 13 or 14
Induction: Either question 13 or 14
Probability + Perms and combs: MC + question 12 simultaneously
Projectile: Question 13
Binomial: Question 14
Parametric: Question 12 or 13
Newtons Method: Question 12

lol every single one of those predictions were in the test

moonbear129 · Oct 26, 2015

probability was one of the last questions in 4unit this year, so didn't think it'll come out agin in 3unit..nek minnit last question..LOL couldn't get last part of last question..anyone knew how to do it?

kawaiipotato · Oct 26, 2015

moonbear129 said:
probability was one of the last questions in 4unit this year, so didn't think it'll come out agin in 3unit..nek minnit last question..LOL couldn't get last part of last question..anyone knew how to do it?

There was no probability in the 4u paper besides the multiple choice one.

moonbear129 · Oct 26, 2015

kawaiipotato said:
There was no probability in the 4u paper besides the multiple choice one.

yes it was..question 16a as well as MC

InteGrand · Oct 26, 2015

moonbear129 said:
probability was one of the last questions in 4unit this year, so didn't think it'll come out agin in 3unit..nek minnit last question..LOL couldn't get last part of last question..anyone knew how to do it?

$The probability $A$ wins is $\frac{1}{2}$ by symmetry.$

$Alternatively, $A$ can win with $B$ winning exactly $k$ games in the first $n+k$ games, where $k$ can be $0,1,2,\ldots,n$ (this happens with probability $\binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k}$; note $B$ can't win \textsl{more} than $n$ games, since then $B$ would win, not $A$), and $A$ winning the last game (with probability $\frac{1}{2}$). Hence probability of $A$ winning with exactly $k$ wins for $B$ is clearly $\binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k}\times \frac{1}{2}$. Summing up over all possible $k$ gives us the probability $A$ wins: $\sum_{k=0}^n \binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k}\times \frac{1}{2}$.$

$Since the chance of $A$ winning is $\frac{1}{2}$, we have$

$\begin{align*} \sum_{k=0}^n \binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k}\times \frac{1}{2}&=\frac{1}{2}\\ \Rightarrow \sum_{k=0}^n \binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k} &= 1 \\ \Rightarrow \sum_{k=0}^n \binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k} \times 2^{2n} &= 2^{2n}\\ \Rightarrow \sum_{k=0}^n \binom{n+k}{n}\cdot 2^{n-k} &= 2^{2n} \quad \Big{(}\text{index laws: } \left(\frac{1}{2}\right)^{n+k} \times 2^{2n} = 2^{n-k}\Big{)}\\ \end{align*}$

$which is precisely what we wanted to prove. $\blacksquare$$

WillSheller · Oct 26, 2015

How easy was the projectile motion? I was expecting a much harder question

kawaiipotato · Oct 26, 2015

moonbear129 said:
yes it was..question 16a as well as MC

Oh oops forgot lol. And the one in 3u this year was done by following a similar pattern from (i) and (ii) where you replace 7 games with n+k games and sum over all possible amount of games (where k = 1,2,...,k+1) as shown above by InteGrand.

InteGrand · Oct 26, 2015

kawaiipotato said:
There was no probability in the 4u paper besides the multiple choice one.

There was a probability Q. about putting counters into cells or columns of a large grid.

Extension 1 predictions? (2 Viewers)

malcolm21

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iforgotmyname

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Crisium

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photastic

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malcolm21

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kawaiipotato

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InteGrand

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WillSheller

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