Three digits are randomly selected without replacement for one 1, two 9's and one 6. What is the possibility that the three digit number is a multiple of 3?
To find the probability that the three-digit number is a multiple of 3, we first need to determine the total number of possible outcomes.
Since we are selecting three digits from a set of four options (1, 9, 9, and 6) without replacement, the total number of possible outcomes can be calculated using the formula for combinations:
Total number of outcomes = C(4, 3) = 4!/[(4-3)! * 3!] = 4
Next, we need to calculate the number of favorable outcomes, i.e., the number of three-digit numbers that are multiples of 3.
A number is divisible by 3 if the sum of its digits is also divisible by 3. Let's examine the possible combinations of digits:
1, 9, 9: Sum of digits = 1 + 9 + 9 = 19 (not divisible by 3)
1, 9, 6: Sum of digits = 1 + 9 + 6 = 16 (not divisible by 3)
9, 9, 6: Sum of digits = 9 + 9 + 6 = 24 (divisible by 3)
Since only one combination results in a sum divisible by 3, the number of favorable outcomes is 1.
Now, we can calculate the probability:
Probability = Favorable outcomes / Total outcomes
Probability = 1/4 = 0.25
Therefore, the probability that the three-digit number selected is a multiple of 3 is 0.25 or 25%.