Alocker combination has three nonzero digits, and digits cannot be repeated. if the first digit is odd, what is the probability that the second digit is even?
a. 1/2
b. 4/9
c. 1/8
d. 1/9

Option A.

Step-by-step explanation:

A locker combination has three nonzero, non-repeated digits.

Total possibilities for the first place would be = 9 digits

Total possibilities for the second place would be = 8 digits [out of 9 no. 1 number is chosen]

and for the third place the possibilities of digits = 7 digits.

If the first digit is odd.

For the second place the even numbers are = 2, 4, 6, and 8

So the probability that the second digit is even = =

Option A. is the answer.

i have no clue at all sorry

step-by-step explanation: