Probability

Jill uses a combination lock on her locker that has $3$ wheels, each labeled with $10$ digits from $0$ to $9$. The combination is a particular sequence with no digits repeating.

1. What is the probability of someone guessing the correct combination?
2. If the digits can be repeated, what are the odds against someone guessing the combination?

1. Let $A$ be the event of guessing the correct combination when digits are not repeated
   
   Since no digits are repeated,
   
   the first digit can be selected (from the digits $0$ to $9$) in $10$ ways;
   
   the second digit can be selected in $9$ ways;
   
   and the third digit can be selected in $8$ ways.
   
   $\therefore \;$ Number of ways of selecting the $3$ digits $= 10 \times 9 \times 8 = 720$ ways.
   
   $\therefore \;$ Number of elements in sample space $S = n\left(S\right) = 720$
   
   The correct combination can be selected in $1$ way.
   
   $\therefore \;$ Number of elements in event $A = n \left(A\right) = 1$
   
   $\therefore \;$ Probability of guessing the correct combination $= P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{1}{720}$
   
   
2. Let $B$ be the event of guessing the correct combination when the digits are repeated
   
   Then, the first digit can be selected (from the digits $0$ to $9$) in $10$ ways;
   
   the second digit can also be selected in $10$ ways;
   
   and the third digit can also be selected in $10$ ways.
   
   $\therefore \;$ Number of ways of selecting the $3$ digits $= 10 \times 10 \times 10 = 1000$ ways
   
   $\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 1000$
   
   The correct combination can only be selected in $1$ way.
   
   $\therefore \;$ Number of elements in event $B = n \left(B\right) = 1$
   
   Let $P \left(s\right)$ be the probability of success of event $B$ and $P \left(f\right)$ be the probability of failure of event $B$
   
   $P \left(s\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{1}{1000}$
   
   $P \left(f\right) = 1- \dfrac{1}{1000} = \dfrac{999}{1000}$
   
   $\therefore \;$ Odds against guessing the correct combination
   
   $= \dfrac{P \left(f\right)}{P \left(s\right)} = \dfrac{999 / 1000}{1 / 1000} = \dfrac{999}{1}$