a padlock has a four - digit code that includes digits from 0 to 9, inclusive. what is the probability that…

a padlock has a four - digit code that includes digits from 0 to 9, inclusive. what is the probability that the code does not consist of all odd digits if the same digit is not used more than once in the code? 120 out of 5,040 120 out of 3,024 2,904 out of 3,024 4,920 out of 5,040

a padlock has a four - digit code that includes digits from 0 to 9, inclusive. what is the probability that the code does not consist of all odd digits if the same digit is not used more than once in the code? 120 out of 5,040 120 out of 3,024 2,904 out of 3,024 4,920 out of 5,040

Answer

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 from 0 - 9) and (r=4), so (P(10,4)=\frac{10!}{(10 - 4)!}=\frac{10!}{6!}=10\times9\times8\times7 = 5040).

Step2: Calculate number of four - digit codes with all odd digits

There are 5 odd digits (1, 3, 5, 7, 9). Using the permutation formula with (n = 5) and (r = 4), we get (P(5,4)=\frac{5!}{(5 - 4)!}=\frac{5!}{1!}=5\times4\times3\times2=120).

Step3: Calculate number of four - digit codes that are not all odd

The number of four - digit codes that are not all odd is (5040-120 = 4920).

Step4: Calculate the probability

The probability that the code does not consist of all odd digits is (\frac{4920}{5040}).

Answer:

4,920 out of 5,040