Dhaka Higher Secondary 2011/7

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/7

Unread post by BdMO » Fri Jan 28, 2011 10:30 pm

In a game Arjun has to throw a bow towards a target and then Karna has to throw a bow toeards the target. One who hits the target first wins. The game continues with Karna trying after Arjun and Arjun trying after Karna until someone wins. The probability of Arjun hitting the target with a single shot is $\frac{2}{5}$ and the probability that Arjun will win the game is the same as that of Karna. What is the probability of Karna hitting the target with a single shot.

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TIUrmi
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Re: Dhaka Higher Secondary 2011/7

Unread post by TIUrmi » Sun Jan 30, 2011 1:22 pm

My approach was pretty much like this:

If Arjun wins in $n + 1$ shots the probability of his winning is = $\left( \frac{3}{5} \right)^n\frac{2}{5}$
If Karna wins while Arjun will fail the probality is = $\frac{3}{5} \left( \frac{x-p}{x} \right)^n\frac{p}{x}$( where p = probability of karan winning)

So,$\left( \frac{3}{5} \right)^n\frac{2}{5}= \frac{3}{5}\frac{x-p}{x}^n\frac{p}{x}$
$\left( \frac{3}{5} \right)^{n-1}\frac{2}{5} = \left( \frac{x-p}{x} \right) ^n\frac{p}{x}$
$\frac{{3^{n-1}\times 2}}{25} =\frac{{p(x-p)}^n}{x^{n+1}}$

From here we can tell x = 5 and p = 2
The probability of Karna hitting the target once and winning is therefore $\frac{3}{5}\frac{2}{5} = \frac{6}{25}$. Though the english statement is 'the probability of Karna hitting the target with a single shot', in Bangla probably it was " ekbar teer chhurey jetar sombhabona koto" or maybe unfortunately I read the statement wrong.

BTW, I am not sure about the solution. If the statement was indeed what I thought ( by mistake) is the solution correct?
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abir91
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Re: Dhaka Higher Secondary 2011/7

Unread post by abir91 » Sun Jan 30, 2011 1:52 pm

I think, either way (whichever understanding of the question we assume), your solution is wrong although I believe you can fix it.

If you are really stuck where is the mistake, here is a pointer:
To calculate the probability of Arjun winning in n + 1 shots, you must take into account that Karna will miss n shots as well.
Abir

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bristy1588
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Re: Dhaka Higher Secondary 2011/7

Unread post by bristy1588 » Tue Nov 29, 2011 2:13 am

Abir Vai,

Is the answer 2/3?
Bristy Sikder

sourav das
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Re: Dhaka Higher Secondary 2011/7

Unread post by sourav das » Tue Nov 29, 2011 12:26 pm

I think so.
My solution:
The probability of wining for Arjun in 2n+1 shots=The probability of wining for Karna in 2n+2 shots
Now,The probability of wining for Arjun in 2n+1 shots = $(\frac{3}{5})^n * \frac{2}{5} * (1-x)^n$ [x is the probability for karna to win in one shot]
The probability of wining for Karna in 2n+2 shots=$(\frac{3}{5})^{n+1} * x * (1-x)^n$
Which implies $x =\frac{2}{3} $
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sm.joty
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Re: Dhaka Higher Secondary 2011/7

Unread post by sm.joty » Thu Dec 29, 2011 3:19 pm

কিছুই বুঝলাম না। :(
কেউ একটু বাংলায় বলেন। আর গানিতিক অপারেশন গুলা একটু ভেঙে বলেন।
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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