Question

a bicycle lock has a four - digit code. the possible digits, 0 through 9, cannot be repeated. what is the probability that the lock code will begin with the number 5? what is the probability that the lock code will not contain the number 0?

Explanation:

Step1: Calculate total number of four - digit codes

The number of permutations of \(n\) distinct objects taken \(r\) at a time is given by \(P(n,r)=\frac{n!}{(n - r)!}\). Here, \(n = 10\) (digits 0 - 9) and \(r=4\), so the total number of four - digit codes with no repeated digits is \(P(10,4)=\frac{10!}{(10 - 4)!}=\frac{10!}{6!}=10\times9\times8\times7 = 5040\).

Step2: Calculate the number of codes starting with 5

If the code starts with 5, then we need to arrange the remaining 9 digits in the remaining 3 positions. So the number of such codes is \(P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7=504\).
The probability that the lock code begins with 5 is \(\frac{504}{5040}=\frac{1}{10}\).

Step3: Calculate the number of codes without 0

If the code does not contain 0, then \(n = 9\) (digits 1 - 9) and \(r = 4\). The number of such codes is \(P(9,4)=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6=3024\).
The probability that the lock code does not contain 0 is \(\frac{3024}{5040}=\frac{3}{5}\).

Answer:

The probability that the lock code will begin with the number 5 is \(\frac{1}{10}\).
The probability that the lock code will not contain the number 0 is \(\frac{3}{5}\).