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A locker 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?

A1/2
B4/9
C1/8
D1/9

Verified Answer :

Answer:

\(\frac{1}{2}\) 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 = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Option A. \(\frac{1}{2}\) is the answer.

Answer:

the answer is actualy b

Step-by-step explanation:

i tried the other answer from the other guy and it was incorrect

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