Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose two doors are locked using digital locker and they have different combination.There are 10 possible combination that might be a combination of either of them.

The first door's looker will block user to access after 4 continuous fail attempts. What is the probability that user fails to open the door?

In second looker the user can try unlimited times but after each fails attempt user should wait 3k seconds .What is the average time it takes a user to open this looker?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

First Locker: Presumably we do not try the same combination twice! The probability our first attempt fails is $\frac{9}{10}$. Given that the first attempt failed, the probability the second attempt fails is $\frac{8}{9}$. Given that the first two attempts failed, the probability the third attempt fails is $\frac{7}{8}$. And given $3$ failures in a row, the probability of a fourth is $\frac{6}{7}$. Multiply. Lots of cancellation, we get $\frac{6}{10}$.

The cancellation is trying to tell us that we missed a simpler argument. And we certainly did. Imagine listing the $10$ combinations at random in a row: we will try the first $4$ numbers in the list. Then we succeed precisely if the combination is among the first $4$ in the list. So the probability we succeed is $\frac{4}{10}$, and therefore the probability we fail is $\frac{6}{10}$. Or equivalently we fail if the combination is among the last $6$ in the list. The probability of this is $\frac{6}{10}$.

Second Locker: Let the mandatory waiting time between tries be $a$ units. If we succeed immediately, our waiting time was $0$. If we succeed on the second try, our total waiting time was $a$. If we succeed on the third try, our total waiting time was $2a$. Continue. Finally, if we succeed on the $10$-th try, our total waiting time was $9a$.

For any trial, the probability we succeed on that trial is $\frac{1}{10}$. This is because the right combination is equally likely to be at any place in our sequence of choices. Thus the expected total waiting time is $$\frac{1}{10}(0)+\frac{1}{10}(a)+\cdots+\frac{1}{10}(9a).$$

This simplifies to $\frac{a}{10}(1+2+\cdots+9)$, which is $\frac{9a}{2}$.

share|improve this answer
    
It is a TeX typo, left out the dollar sign. Was fixed a while ago. –  André Nicolas Nov 22 '12 at 18:41
    
Thanks , Really a good answer, Thanks –  Hooman Nov 22 '12 at 18:42
    
I showed the calculation of mean time. Recall that if $X$ takes on values $x_1,x_2,\dots,x_n$ with probabilities $p_1,p_2,\dots,p_n$ then $E(X)=x_1p_1+x_2p_2+\cdots+x_np_n$. In this case we could use also use symmetry to find the answer, but I wanted to do it in a straightforward "unclever" way. –  André Nicolas Nov 22 '12 at 19:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.