Moon vai iz gonna to hack my acount :x :x

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Hasib
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Moon vai iz gonna to hack my acount :x :x

Unread post by Hasib » Mon Jan 31, 2011 3:00 pm

Moon vai iz my enemy. He is gonna to hack my forum's account! :x :x :x Grrrr
Somehow, he knows that my password has 10 digits and all of them are numbers. He can input password 25 times. What is the probability of hacking my account by Moon vai my enemy? :P :P
A man is not finished when he's defeated, he's finished when he quits.

tanvirab
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by tanvirab » Mon Jan 31, 2011 11:12 pm

মুনকে দোররা!

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Zzzz
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Zzzz » Tue Feb 01, 2011 9:19 am

মুন ভাইয়ার দোররা খাওয়ার সফলতার সম্ভাবনা দেখা যাচ্ছে খুবই কম... $\frac {25} {10^{10}}$
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Hasib
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Hasib » Tue Feb 01, 2011 4:10 pm

Hehehehe
when he input first time, his probabilty of success is $\frac{1}{10^{10}}$ But, if he unsucced, he will input for $2^{nd}$ time. This time, he won't input that password which one he input first time. So, this times probability is $\frac{1}{10^{10}-1}$

now, another abuse, iff he unsucced in $n^{th}$ then he will try for $(n+1)^{th}$ time. So, can i add all the possibility for $k^{th}\text{ [whence} 1\le k \le 25]$ time?


I can't fix the problem yet :( :cry: only 10 days for BdMO, but i am so dull yet. :cry: what i will do in BdMO? :?
A man is not finished when he's defeated, he's finished when he quits.

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Tahmid Hasan
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Tahmid Hasan » Tue Feb 01, 2011 6:44 pm

i think ur ans is wrong hasib :(
বড় ভালবাসি তোমায়,মা

Hasib
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Hasib » Tue Feb 01, 2011 6:54 pm

i havent give any solution! Where is my solution? I just say we cant add $\frac{1}{10^{10}}+\frac{1}{10^{10}}+\frac{1}{10^{10}}.......$

cause, for 2nd trial he wont input that one which he already inputed 1st trial. For 3rd trial he wont input that ones which he inputed 2nd and 1st trial....



Can u get it?
A man is not finished when he's defeated, he's finished when he quits.

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Tahmid Hasan
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Tahmid Hasan » Tue Feb 01, 2011 7:14 pm

the ans is the same as others sry
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HandaramTheGreat
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by HandaramTheGreat » Wed Feb 02, 2011 1:20 pm

probability of hacking account = probability of first time + second time+ ... + $25^{th}$ time
probability of success in first time = $\frac{1}{10^{10}}$
probability of success in second time
=probability of not being succeeded in first time $\times$ probability of success in second time
=$\frac{10^{10}-1}{10^{10}} \cdot \frac{1}{10^{10}-1}$ (moon vai so clever that he wouldn't enter a password twice :P )
=$\frac{1}{10^{10}}$
probability of success in third time
=prob. of not being succeeded is first time $\times$ prob. of not in second time $\times$ prob. of success in third time
=$\frac{10^{10}-1}{10^{10}} \cdot \frac{10^{10}-2}{10^{10}-1} \cdot \frac{1}{10^{10}-2}$
=$\frac{1}{10^{10}}$
.
.
.
sum of all =$\frac{25}{10^{10}}$

আমার মস্তিষ্ক কি সুস্থ আছে? কেউ একজন নিশ্চিত করলে শান্তি পাই...

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Zzzz
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by Zzzz » Wed Feb 02, 2011 1:51 pm

হুম, আমারও তো তাই আসলো :roll:
Every logical solution to a problem has its own beauty.
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HandaramTheGreat
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Re: Moon vai iz gonna to hack my acount :x :x

Unread post by HandaramTheGreat » Wed Feb 02, 2011 4:19 pm

আমি আসলে জানতে চাচ্ছিলাম আমার পদ্ধতি/যুক্তিটা ঠিক আছে কিনা...
hasib.mo wrote:... we cant add $\frac{1}{10^{10}}+\frac{1}{10^{10}}+\frac{1}{10^{10}}.......$
hasib thought that you did it in that way...

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