A locker can be opened by dialing a fixed thr | vedclass

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(B) The total number of possible codes from $000$ to $999$ is $1000$.Let $E_i$ be the event that the stranger fails at the $i^{th}$ trial.The probabil

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$\frac{k}{1000}$

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(B) The total number of possible codes from $000$ to $999$ is $1000$.Let $E_i$ be the event that the stranger fails at the $i^{th}$ trial.The probability of failing at the first trial is $P(E_1) = \frac{999}{1000}$.If the stranger fails at the first trial,there are $999$ codes remaining,and $998$ of them are incorrect. Thus,the probability of failing at the second trial given failure at the first is $P(E_2|E_1) = \frac{998}{999}$.For the stranger to succeed at the $k^{th}$ trial,they must fail the first $(k-1)$ trials and then succeed on the $k^{th}$ trial.The probability of failing the first $(k-1)$ trials is $P(E_1 \cap E_2 \cap ... \cap E_{k-1}) = \frac{999}{1000} 	imes \frac{998}{999} 	imes ... 	imes \frac{1000-(k-1)}{1000-(k-2)} = \frac{1000-k+1}{1000}$.Given that the stranger has failed $(k-1)$ times,there are $1000-(k-1) = 1001-k$ codes remaining,one of which is the correct code.The probability of succeeding on the $k^{th}$ trial given failure in the previous $(k-1)$ trials is $\frac{1}{1001-k}$.Therefore,the probability of success at the $k^{th}$ trial is $\frac{1000-k+1}{1000} 	imes \frac{1}{1001-k} = \frac{1}{1000}$.

$A$ locker can be opened by dialing a fixed three-digit code (between $000$ and $999$). $A$ stranger who does not know the code tries to open the locker by dialing three digits at random. The probability that the stranger succeeds at the $k^{th}$ trial is

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