Example — Safe

Example — Safe, patient burglar

A safe has a combination of $5$ digits from $0$ to $9$.

(a) The probability of guessing it on the first attempt is $1/10^5 = 1/100\,000$: the total cases are the arrangements with repetition, that is $10^5$.

(b) If $1000$ attempts are made, all different, the probability of getting it at least once is computed with a combinatorial argument. The total cases are the ways of choosing $1000$ distinct attempts out of $10^5$, that is $\binom{10^5}{1000}$. The favourable cases are those in which the correct combination is among the $1000$ chosen attempts, that is $\binom{10^5-1}{1000-1}$. Simplifying:

$P = \frac{\binom{10^5-1}{10^3-1}}{\binom{10^5}{10^3}} = \frac{10^3}{10^5} = \frac{1}{100}.$

Links

Topics: Probability
Concepts: Simple combinations · Arrangements with repetition · Probability
Methods: Simple combinations · Arrangements with repetition
Skills: Combinatorics · Probability calculation