A padlock has a four-fidget 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? Option 1: 120 out of 5,040 Option 2: 120 out of 3,024 Option 3: 2,904 out of 3,024 Option 4: 4,920 out of 5,040

option 4

Step-by-step explanation:

Probability = number of favourable outcomes / total number of possible outcomes

total number of codes with an all odd digit = 3 x 5 x 7 x 2 = 120

total number of possible outcomes = 10 x 9 x 8 x 7 = 5040

probability all digits are odd = 120 / 5040

probability that the code does not consist of all odd digits = 1 - probability all digits are odd

1 - 120 / 5040 = 4920 / 5040

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step-by-step explanation:

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

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