Author Topic: HSC 2011 Q5a (Read 826 times) Tweet Share

MinterMints · « **on:** October 09, 2018, 08:38:18 pm »

Just a bit stuck on part (iv), don't understand the provided solutions.

fun_jirachi · « **Reply #1 on:** October 09, 2018, 09:18:20 pm »

MinterMints · « **Reply #2 on:** October 10, 2018, 01:56:33 pm »

Thanks for the help!
Could you guys also explain this question to me, ive run into them previously, but im not too sure where to start!

RuiAce · « **Reply #3 on:** October 10, 2018, 05:05:47 pm »

Quote from: MinterMints on October 10, 2018, 01:56:33 pm

Thanks for the help!
Could you guys also explain this question to me, ive run into them previously, but im not too sure where to start!

$\text{Using binomial probabilities, the probability he hits}\\ \text{the targets once or twice, given 2 tries, is}\\ \binom{2}{2}p^2 + \binom{2}{1} p(1-p).$
$\text{i.e. }p^2 + 2p(1-p)\text{, which expands and simplifies to }2p-p^2.$
Alternatively, the complement event is that he misses on both shots, and thus the probability is $ 1 - (1-p)^2$. This also expands and simplifies to $2p-p^2$.
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$\text{Again via binomial probabilities, the probability he hits}\\ \text{the target twice or three times, given 3 tries, is}\\ \binom33 p^3 +\binom32 p^2(1-p).$
$\text{i.e. }p^3 + 3p^2(1-p)\text{, which expands and simplifies to }3p^2 - 2p^3$
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$\text{This part requires us to prove that the expression in i)}\\ \text{is greater than the expression in ii), i.e. }\boxed{2p-p^2 > 3p^2 -2 p^3}.$
$\text{We can prove this by proving that }LHS - RHS > 0.\\ \text{In our case,}\\ \begin{align*}LHS - RHS &= (2p-p^2) - (3p^2-2p^3)\\ &= 2p - 4p^2 + p^3\\ &= 2p (p^2-2p+1)\\ &= 2p(p-1)^2\end{align*}$
$\text{But then, we are essentially told that }p > 0\\ \text{which immediately implies that }2p>0.\\ \text{Also, since }p\neq 1, \, (p-1)^2 > 0.\\ \text{Hence upon multiplying these inequalities, }2p(1-p)^2 > 0.\\ \therefore LHS - RHS > 0\text{ as required.}$
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$\text{Whereas in iv), we are not interested in just the inequality.}\\ \text{We want the specific value of }p\text{ that ensures that}\\ \boxed{2p-p^2 = 2(3p^2-2p^3)}.$
$\text{Solving, noting }p\neq 0,\\ \begin{align*} 2p-p^2 &= 6p^2 - 4p^3\\ 2 - p &= 6p - 4p^2\\ 4p^2 - 7p + 2 &= 0\\ p&= \frac{7\pm \sqrt{49 - 32}}{8}\\ &= \frac{7\pm\sqrt{17}}{8}. \end{align*}$
$\text{According to the calculator, the only solution satisfying }0<p<1\\ \text{is }p = \frac{7-\sqrt{17}}{8}$

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Author Topic: HSC 2011 Q5a (Read 826 times) Tweet Share

MinterMints

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