Page 1 of 1

Dhaka Higher Secondary 2011/7

Posted: Fri Jan 28, 2011 10:30 pm
by BdMO
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

Posted: Sun Jan 30, 2011 1:22 pm
by TIUrmi
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?

Re: Dhaka Higher Secondary 2011/7

Posted: Sun Jan 30, 2011 1:52 pm
by abir91
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.

Re: Dhaka Higher Secondary 2011/7

Posted: Tue Nov 29, 2011 2:13 am
by bristy1588
Abir Vai,

Is the answer 2/3?

Re: Dhaka Higher Secondary 2011/7

Posted: Tue Nov 29, 2011 12:26 pm
by sourav das
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} $
If you can find any bug, please inform me.

Re: Dhaka Higher Secondary 2011/7

Posted: Thu Dec 29, 2011 3:19 pm
by sm.joty
কিছুই বুঝলাম না। :(
কেউ একটু বাংলায় বলেন। আর গানিতিক অপারেশন গুলা একটু ভেঙে বলেন।