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?
A. 1/2
B. 4/9
C. 1/8
D. 1/9

Answer :

Answer:

[tex]\frac{1}{2}[/tex] 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 = [tex]\frac{4}{8}[/tex] = [tex]\frac{1}{2}[/tex]

Option A. [tex]\frac{1}{2}[/tex] 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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