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

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

yap! I thought zubaer vai did $\frac{1}{10^{10}}\times 25$

but, i was wrong! But, lets me know, why $\text{2nd times success probability}=\text{1st times unsuccess probability}\times \text{ 2nd time's success probability}$?
A man is not finished when he's defeated, he's finished when he quits.

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

hasib.mo wrote:yap! I thought zubaer vai did $\frac{1}{10^{10}}\times 25$

but, i was wrong! ...
Hey.. I actually did it

Look, think about this problem: there are $5$ vowels, what is the probability that if I pick one, it will be $a$?
Answer is obvious (though its a dangerous word in mathematics ) $\frac {1}{5}$

Now, there are $4!$ strings that starts with $a$ and $5$ vowels can make $5!$ strings. So the probability of picking a string which starts with $a$ is $\frac {4!}{5!}= \frac {1}{5}$ the same ! but why?
probability of picking $a=\frac {1}{5}=\frac{1\cdot 4!}{5\cdot 4!}$

To convert to strings, we just need to multiply the numerator and denominator by $4!$, so probability remains same.

In Moon vai's problem, there are $10^{10}$ possible passwords. Moon vai can try these passwords in $10^{10}!$ orders (forget $25$). Now he will succeed if he tries an order which contains the right password before $26^{th}$ place. How many orders of password contains the right one in $1^{st}$ place? Answer = $(10^{10}-1)!$ So, what is the probability that he will get it in first attempt? Answer= $\frac {(10^{10}-1)!}{10^{10}!}=\frac {1}{10^{10}}$. Similarly how many orders contain right one in second place? Answer= $(10^{10}-1)!$ The probability that he will get it in second attempt? Answer= $\frac {(10^{10}-1)!}{10^{10}!}= \frac {1}{10^{10}}$.....

The answer= $25 \times \frac {1}{10^{10}}$
Every logical solution to a problem has its own beauty.