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14.Probability
hard
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
A
$\frac{k}{{999}}$
B
$\frac{k}{{1000}}$
C
$\frac{{k - 1}}{{1000}}$
D
None of these
Solution
(b) 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}}$
$⇒$ $P(A) = 1 – \frac{{1000 – k}}{{1000}} = \frac{k}{{1000}}.$
Standard 11
Mathematics
