A) \[\frac{k}{999}\]
B) \[\frac{k}{1000}\]
C) \[\frac{k-1}{1000}\]
D) None of these
Correct Answer: B
Solution :
Let A denote the event that the stranger succeeds at the \[{{k}^{th}}\] trial. Then \[P(A')=\frac{999}{1000}\times \frac{998}{999}\,\,\times .....\times \] \[\frac{1000-k+1}{1000-k+2}\times \frac{1000-k}{1000-k+1}\] \[\Rightarrow \,\,\,P(A')=\frac{1000-k}{1000}\Rightarrow P(A)=1-\frac{1000-k}{1000}=\frac{k}{1000}\]You need to login to perform this action.
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