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 \[P({A}')=\frac{999}{1000}\times \frac{998}{999}\times .....\times \frac{1000-k+1}{1000-k+2}\times \frac{1000-k}{1000-k+1}\] \[\Rightarrow \]\[P({A}')\]\[=\frac{1000-k}{1000}\] Þ \[P(A)=1-\frac{1000-k}{1000}=\frac{k}{1000}.\]
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