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? 0.1. what is the probability that the lock code will not contain the number 0? 0.1 0.5 0.6 0.8

Explanation:

Step1: Calculate total permutations

The total number of 4 - digit codes with non - repeated digits from 0 - 9 is given by the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 10\) (digits 0 - 9) and \(r=4\). So \(P(10,4)=\frac{10!}{(10 - 4)!}=\frac{10!}{6!}=10\times9\times8\times7 = 5040\).

Step2: Calculate permutations without 0

If we exclude 0, we have 9 digits (1 - 9) to form a 4 - digit code with non - repeated digits. Using the permutation formula with \(n = 9\) and \(r = 4\), we get \(P(9,4)=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6=3024\).

Step3: Calculate the probability

The probability \(P\) that the lock code does not contain 0 is the ratio of the number of favorable permutations (without 0) to the total number of permutations. So \(P=\frac{P(9,4)}{P(10,4)}=\frac{3024}{5040}=\frac{9\times8\times7\times6}{10\times9\times8\times7}=\frac{6}{10}=0.6\)? Wait, no, wait. Wait, let's recalculate. Wait, \(P(9,4)=9\times8\times7\times6 = 3024\), \(P(10,4)=10\times9\times8\times7 = 5040\). Then \(\frac{3024}{5040}=\frac{9\times8\times7\times6}{10\times9\times8\times7}=\frac{6}{10}\)? No, wait \(9\times8\times7\times6=3024\), \(10\times9\times8\times7 = 5040\), \(3024\div5040=\frac{3024}{5040}\). Divide numerator and denominator by 504: \(3024\div504 = 6\), \(5040\div504 = 10\), so \(\frac{6}{10}=0.6\)? Wait, but let's think in terms of probability step - by - step for each digit.

For the first digit, the probability of not choosing 0 is \(\frac{9}{10}\) (since there are 9 non - zero digits out of 10). For the second digit, since we have already chosen one non - zero digit, there are 8 non - zero digits left out of 9 remaining digits, so the probability is \(\frac{8}{9}\). For the third digit, there are 7 non - zero digits left out of 8 remaining digits, probability \(\frac{7}{8}\). For the fourth digit, there are 6 non - zero digits left out of 7 remaining digits, probability \(\frac{6}{7}\).

Then the total probability is \(\frac{9}{10}\times\frac{8}{9}\times\frac{7}{8}\times\frac{6}{7}\). The 9 in the numerator and denominator cancels, the 8 in the numerator and denominator cancels, the 7 in the numerator and denominator cancels, and we are left with \(\frac{6}{10}=0.6\).

Answer:

0.6