Probability and graph question (1 Viewer)

Jonothancroller · Aug 17, 2016

Hey guys, I need help with these two questions: http://m.imgur.com/a/39Qms
Any help would be greatly appreciated, thanks.

InteGrand · Aug 17, 2016

Jonothancroller said:
Hey guys, I need help with these two questions: http://m.imgur.com/a/39Qms
Any help would be greatly appreciated, thanks.

Quick way for the first one: in the extreme case where n = N (i.e. everyone in the pool is a James Ruse student), we are guaranteed that the next student picked is from JR, so the probability would be 1. The only option that is 1 when n = N is option (A).

Thus the answer is (A).

Jonothancroller · Aug 17, 2016

How would you get a general solution for a question like that though? I must admit I never thought of that method though, thanks for that!

InteGrand · Aug 17, 2016

$\noindent For the second one, note that if $k=0$, the polynomial is $P(x) = x^3 -8$, which doesn't have only real roots (it factors as $\left(x-2\right)\left(x^2 + 2x + 4\right)$, and this quadratic has no real roots, so $P(x)$ has some non-real roots. Alternatively, if you know that cubics have three complex roots in total, you'll know this cubic has two non-real roots, since it clearly has only one real root.). This eliminates options (A) and (B).$

$\noindent Now suppose $k=2$, then $P(x) = x^3 -2x^2 + 4x -8$. Thus$

$\begin{align*}P(x) &= x^2 \left(x-2\right) +4 \left(x-2\right) \\ &= \left(x^2 + 4\right) (x-2), \end{align*}$

$\noindent and again this quadratic factor has no real roots. This eliminates option (C). Hence the answer is (D).$

InteGrand · Aug 17, 2016

Jonothancroller said:
How would you get a general solution for a question like that though? I must admit I never thought of that method though, thanks for that!

$\noindent Note that we need to assume $n\geq 1$ and $N \geq 2$ for this question to make sense (since we picked two students so the pool size is at least 2, and we know there is at least one JR student, since we are told we picked one).$

$\noindent The total no. of ways to pick two students is $A \equiv \binom{N}{2}$. The total no. of ways $B$ to pick at least one JR student can be found by subtracting $X$ from $A$, where $X$ is the no. of ways to pick two students where none are JR ones. So $X = \binom{N-n}{2}\Rightarrow B = A -X = \binom{N}{2} -\binom{N-n}{2} = \frac{N\left(N-1\right)}{2} - \frac{\left(N-n\right)\left(N-n-1\right)}{2}$. The total no. of ways $C$ in which both students are JR ones is just $\binom{n}{2} = \frac{n(n-1)}{2}$ (by convention this binomial coefficient is $0$ if $n<2$). So the desired probability is $\frac{C}{B}$. If you simplify this, it should lead to the desired answer.$

$\noindent In general after obtaining answers to Q's like this where you come up with some sort of formula, it's a good idea to test your obtained formula by seeing if it holds in easy-to-test extreme and simple cases where we know what the answer should be (e.g. we expect our formula to yield $1$ when $n=N$; we expect it to yield $0$ if $n=1$, since if there's only one JR student, there's no chance of picking a second one; etc.). The good thing about checks like this is that they are easy to do and if any of them fail, it tells us we've made a mistake somewhere.$

Jonothancroller · Aug 17, 2016

Thanks so much! That was really helpful.

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Probability and graph question (1 Viewer)

Jonothancroller

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