Question

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 1

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 2

_ _ _ here are the three digits

Total:
5 ways to choose the first (1, 3, 5, 7, 9)
9 for the second
8 for the third
5*8*9=360.

5 ways to choose the first digit (1, 3, 5, 7, 9)
5 ways to choose the second ( 2, 4, 6, 8, 0)
8 for the third
5*5*8=200
200/360=20/36=10/18=5/9, but it's not one of the choices.