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

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(B) There are $1000$ possible codes ranging from $000$ to $999$.Let $E$ be the event that the stranger succeeds at the $k^{th}$ trial.This means the s

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$\frac{k}{1000}$

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(B) There are $1000$ possible codes ranging from $000$ to $999$.Let $E$ be the event that the stranger succeeds at the $k^{th}$ trial.This means the stranger fails in the first $(k-1)$ trials and succeeds in the $k^{th}$ trial.Since the stranger does not know the code,they choose a code at random. Assuming they do not repeat the same code:The probability of failing the first trial is $\frac{999}{1000}$.The probability of failing the second trial given the first failed is $\frac{998}{999}$.The probability of failing the $(k-1)^{th}$ trial is $\frac{1000-(k-1)}{1000-(k-2)}$.The probability of succeeding at the $k^{th}$ trial is $\frac{1}{1000-(k-1)}$.Multiplying these probabilities: $P(E) = \left( \frac{999}{1000} 	imes \frac{998}{999} 	imes \dots 	imes \frac{1000-k+1}{1000-k+2} ight) 	imes \frac{1}{1000-k+1} = \frac{1}{1000}$.However,if the stranger can repeat codes,the probability of success at each trial is independent and equal to $\frac{1}{1000}$.Thus,the probability of succeeding at the $k^{th}$ trial is $\frac{1}{1000}$ if trials are independent. Given the options provided,the intended answer is $\frac{1}{1000}$ (which corresponds to the probability of success on any specific trial if codes are replaced/repeated).

$A$ 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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