A locker combination consists of two non-zero digits, and each combination consists of different digits. Event A is defined as choosing an even number as the first digit, and event B is defined as choosing an even number as the second digit. If a combination is picked at random, with each possible locker combination being equally likely, what is P(A and B) expressed in simplest form? A. 1/6 B. 5/18 C. 1/2 D. 5/9

B. 5/18

Step-by-step explanation:

We have to digits taking the 0, they are as follows:

1, 2, 3, 4, 5, 6, 7, 8, 9

In other words, there are 9 in total.

Odd are 5 (1, 3, 5, 7, 9)

Pairs are 4 (2, 4, 6, 8)

Now, we have that the probability of drawing an odd one would be: 5/9

and that also, already selected that number, there would be 8 in total.

Therefore, drawing another number that is even the probability would be: 4/8

So the final probability is:

(5/9) * (4/8) = 20/72, we simplify (divide by 4) and we have 5/18 left

In other words, the answer is B. 5/18

because they could drop even lower..

ab will be equal to