HELP Binomial theorem exam (1 Viewer)

the_matrix · Jun 11, 2016

Hello, I've got my HSC Task 3 on Tuesday and binomial theorem is in it. I am really struggling with this topic like I did in permutations & combinations. Could anyone please help me answer these 4 questions:

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InteGrand · Jun 11, 2016

the_matrix said:
Hello, I've got my HSC Task 3 on Tuesday and binomial theorem is in it. I am really struggling with this topic like I did in permutations & combinations. Could anyone please help me answer these 4 questions:

<a href="http://tinypic.com?ref=4k6xww" target="_blank"><img src="http://i65.tinypic.com/4k6xww.png" border="0" alt="Image and video hosting by TinyPic"></a>

Here are some hints.

$\noindent Q.7. Recall that $\left(3+4x\right)^n = \sum _{k=0}^{n}\binom{n}{k}3^{n-k}4^k x^k$ from the Binomial Theorem. So the $x^2$ coefficient is (with $k=2$) $\binom{n}{2}3^{n-2}\times 16$ and the $x^3$ coefficient is (with $k=3$) $\binom{n}{3}3^{n-3}\times 64$. We are thus given that $\frac{\binom{n}{2}3^{n-2}\times 16}{\binom{n}{3}3^{n-3}\times 64}=\frac{3}{4}$. From this, we just need to solve for the positive integer $n$. Can you get it from here?$

$\noindent Q.2. Recall that $\left(1+x\right)^n = \sum _{k=0}^{n}\binom{n}{k}x^k$ from the Binomial Theorem. Can you get it from here? Sub. in appropriate $x$ values to both sides of the equation here now. E.g. for (a), sub. $x=1$, and remember to take the $k=0$ term to the other side, since the sum in the question starts at $k=1$.$

$\noindent Q.4. Assume any given child is equally likely to be boy or girl. Then for a given family, the probability they have exactly 2 boys and 2 girls is $p=\binom{4}{2}\left(0.5\right)^2\left(0.5\right)^2$ (can you see why? It's basically just binomial probability formula). Now again from binomial probability, the answer to the question will be $\binom{6}{2}p^{2}\left(1-p\right)^{6-2}$, where $p$ is defined above.$

$\noindent Q.5. This is again binomial probability, since it is sampling with replacement. Since the probability of a red marble being drawn in any given draw is $p=\frac{6}{10}$, the probability that from $6$ draws we get exactly $k$ red marbles is $p_{k} = \binom{6}{k}p^{k}\left(1-p\right)^{6-k}$, for $k=0,1,\ldots,6$ (standard binomial probability). You can use this to find the answers to (a) and (b).$

the_matrix · Jun 14, 2016

Thank you so much for your help! Also, i found out that my working out was pretty much same as yours fr Q 4 and 5, its judt that the answers were wrong.

HELP Binomial theorem exam (1 Viewer)

the_matrix

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InteGrand

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the_matrix

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