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ser-zykov [4K]
3 years ago
14

A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons

ist of all odd digits if the same digit is not used more than once in the code?
Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
5 0
The answer is D: <span>4,920 out of 5,040</span>
Eduardwww [97]3 years ago
4 0
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

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