The digits 5, 6, 7, 8, and 9 are randomly arranged to form a three-digit number. (digits are not repeated) find the probability that the number is even and greater than 900 the probability that the three-digit number is even and greater than 900 is (type an integer or a simplified fraction.)

The probability that the three digit number is even and greater than 900 is 0.1

Step-by-step explanation:

Since digits are not repeated, we have 5 options for the first digit, then 4 options for the second one and 3 options for the third one. The amount of possible 3 digit numbers is: 5 x 4 x 3 = 60.

a) If the number is even and greater than 900, then the first digit has to be 9 (so we have only one option), the third digit can be either 6 or 8 (2 options) and the second digit can be either of the three remaining digits. Therefore the amount of digits possible is: 1 x 3 x 2 = 6.

The probability of the number being even and greater than 900 is 6/60= 0.1

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!