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

For discussing Olympiad Level Combinatorics problems
Hasib
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Moon vai iz gonna to hack my acount :x :x

Moon vai iz my enemy. He is gonna to hack my forum's account! 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?
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

মুনকে দোররা!

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

মুন ভাইয়ার দোররা খাওয়ার সফলতার সম্ভাবনা দেখা যাচ্ছে খুবই কম... $\frac {25} {10^{10}}$
Every logical solution to a problem has its own beauty.

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

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 only 10 days for BdMO, but i am so dull yet. what i will do in BdMO?
A man is not finished when he's defeated, he's finished when he quits.

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

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

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.

Tahmid Hasan
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Joined: Thu Dec 09, 2010 5:34 pm

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

the ans is the same as others sry
বড় ভালবাসি তোমায়,মা

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

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 )
=$\frac{1}{10^{10}}$
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 =$\frac{25}{10^{10}}$

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

হুম, আমারও তো তাই আসলো
Every logical solution to a problem has its own beauty.
hasib.mo wrote:... we cant add $\frac{1}{10^{10}}+\frac{1}{10^{10}}+\frac{1}{10^{10}}.......$