Dhaka Higher Secondary 2011/7
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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.
Re: Dhaka Higher Secondary 2011/7
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?
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?
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter
Re: Dhaka Higher Secondary 2011/7
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:
If you are really stuck where is the mistake, here is a pointer:
- bristy1588
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Re: Dhaka Higher Secondary 2011/7
I think so.
My solution:
If you can find any bug, please inform me.
My solution:
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Dhaka Higher Secondary 2011/7
কিছুই বুঝলাম না।
কেউ একটু বাংলায় বলেন। আর গানিতিক অপারেশন গুলা একটু ভেঙে বলেন।
কেউ একটু বাংলায় বলেন। আর গানিতিক অপারেশন গুলা একটু ভেঙে বলেন।
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