## 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

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.

TIUrmi
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### 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?
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter

abir91
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### 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:
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

Have you read the Forum Rules and Guidelines?

bristy1588
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### Re: Dhaka Higher Secondary 2011/7

Abir Vai,

Bristy Sikder

sourav das
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### Re: Dhaka Higher Secondary 2011/7

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

কিছুই বুঝলাম না।
কেউ একটু বাংলায় বলেন। আর গানিতিক অপারেশন গুলা একটু ভেঙে বলেন।
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