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

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Question

(b) Let $A$ denote the event that the stranger succeeds at the ${k^{th}}$ trial. Then

$P(A') = \frac{{999}}{{1000}} 	imes \frac{{998}}{{999}}

Vedclass · Accepted Answer

$\frac{k}{{1000}}$

Vedclass · Answer

(b) Let $A$ denote the event that the stranger succeeds at the ${k^{th}}$ trial. Then

$P(A') = \frac{{999}}{{1000}} 	imes \frac{{998}}{{999}} 	imes ..... 	imes \frac{{1000 - k + 1}}{{1000 - k + 2}} 	imes \frac{{1000 - k}}{{1000 - k + 1}}$

$ \Rightarrow $$P(A')$$ = \frac{{1000 - k}}{{1000}}$

$&rArr;$ $P(A) = 1 - \frac{{1000 - k}}{{1000}} = \frac{k}{{1000}}.$

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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