HSC 2016 MX1 Marathon (archive) (1 Viewer)

InteGrand · Jan 31, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Are the green marbles all identical ? Also does order matter ?

$\noindent The green marbles are not like indistinguishable particles (think of each green marble having a label with a number from 1 to $n-1$ if you want; but all green marbles are the same basically. Like there's $n-1$ of them, and we can't `see' any difference in them, but there are $n-1$ separate ones, like imagine having a bunch of marbles lol). Whether order matters or not doesn't affect the answer.$

Paradoxica · Jan 31, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION}$

$\noindent Let $n$ and $k$ be positive integers with $k\leqslant n$. Suppose there are a total of $n$ marbles in an urn and exactly one of these is red, and the others are all green. Alice randomly selects $k$ marbles from the urn without replacement. What is the probability (in terms of $n$ and $k$) that one of her selected marbles is the red one?$

$$\noindent The probability that Alice gets a red marble is the complement to the probability that she only draws green marbles. We treat the problem as if Alice continually draws one marble at a time until she draws $k$ marbles. At the initial draw, her probability of getting a green marble is $\frac{n-1}{n}$ and the numerator and denominator both drop by one for each subsequent draw. Her second draw is $\frac{n-2}{n-1}$, third draw is $\frac{n-3}{n-2}$, and so it follows that her $k^{\text{th}}$ draw is $\frac{n-k}{n-k+1}$. Multiplying these probabilities together cancels out all the numerators and denominators of all intermediate terms, leaving only the denominator of the first term, and the numerator of the last term. So the probability of getting only green marbles with $k$ draws is $\frac{n-k}{n}$. Hence, the probability that Alice gets a red marble over $k$ draws is $\frac{k}{n}$

$\noindent For additional verification, we can clearly see that if $k=n$ then the probability is $100\%$, which is true by the pigeonhole principle.$

InteGrand · Jan 31, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$\noindent The probability that Alice gets a red marble is the complement to the probability that she only draws green marbles. We treat the problem as if Alice continually draws one marble at a time until she draws $k$ marbles. At the initial draw, her probability of getting a green marble is $\frac{n-1}{n}$ and the numerator and denominator both drop by one for each subsequent draw. Her second draw is $\frac{n-2}{n-1}$, third draw is $\frac{n-3}{n-2}$, and so it follows that her $k^{\text{th}}$ draw is $\frac{n-k}{n-k+1}$. Multiplying these probabilities together cancels out all the numerators and denominators of all intermediate terms, leaving only the denominator of the first term, and the numerator of the last term. So the probability of getting only green marbles with $k$ draws is $\frac{n-k}{n}$. Hence, the probability that Alice gets a red marble over $k$ draws is $\frac{k}{n}$

$\noindent For additional verification, we can clearly see that if $k=n$ then the probability is $100\%$, which is true by the pigeonhole principle.$

Correct!

$\noindent By the way, for any curious HSC students, here's an easier way that also explains why the answer is so simple in the end (and is probably what your intuition told you it should be).$

$\noindent Visualise $n$ blank spaces: $\_$ $\_$ $\_$ $\ldots$ $\_$.$

$\noindent Imagine randomly arranging the $n-1$ green marbles and the 1 red marble along this line in the blank spaces. Let the first $k$ marbles in this list be the ones Alice picks. Then the probability that Alice picks a red marble is equivalent to the probability that the red marble will fall into one of the first $k$ spots. Since there are $n$ places to put the red marble and $k$ desirable places (the first $k$ in the list), the desired probability is $\frac{k}{n}$.$

Paradoxica · Jan 31, 2016

Re: HSC 2016 3U Marathon

$\noindent There is a room with $k$ people in it. Assume birthdays are random, leap years do not exist, and that twins and other instances of multi-births do not exist. What is the least possible $k$ such that the probability of two people sharing the same birthday exceeds 99.95$\%$?$

Flop21 · Feb 2, 2016

Re: Flop math question thread

How do I solve this trig equation?

tan^2 (x) + tan(x) - 2 = 0 ; 0=<theta=<360

(x) is actually a theta.

I got an answer of tan(x)=2 and -1, but not sure what to do with this answer afterwards. Or if it's even correct.

sida1049 · Feb 2, 2016