Three digits are randomly selected without replacement from 1, 9, 9 and 6. What is the probability that the three digits form a three-digit number that is a multiple of 3?
This means that you cannot choose the 1. You can compute the probability using different methods. In this case there is a very simple method.
To randomly choose 3 numbers you can put the 4 numbers in some random order and then choose the last 3. The probability that the 1 won't be chosen is then the probability that the 1 will be in the first place if you randomly reshuffle the numbers.
The probability that the 1 will be in any given place is equal to 1/4, so the answer is 1/4.
The 1 is the only problem. The others are multiples of three and if added together will also produce a number divisible by three.
Before any numbers are chosen there are 4 numbers, three of which are NOT the number 1. The probability that the first number chosen will NOT be the 1 is 3/4.
After whatever number other than 1 was selected first, there will be a total of 3 numbers left, 2 of which are NOT the number 1. The probability that the second number chosen is NOT the 1 is 2/3.
After whatever number other than 1 was selected second, there will be a total of 2 numbers left, 1 of which is NOT the number 1. The probability that the thrid number chosen is NOT the 1 is 1/2.
The probability that all three of these events wil occur is the product of the individual probabilities.
To find the probability that the three digits form a three-digit number that is a multiple of 3, we need to determine the number of favorable outcomes and the total number of possible outcomes.
First, let's consider the favorable outcomes. A three-digit number is a multiple of 3 if the sum of its digits is also a multiple of 3.
We have four digits to choose from: 1, 9, 9, and 6.
To form a three-digit number, we need to choose three digits. Since each digit can only be selected once (without replacement), we have six possible arrangements of the three digits.
The favorable outcomes are the arrangements that yield a sum of digits that is a multiple of 3. Let's calculate these arrangements:
1. Possible arrangements with a sum of 3: (1, 1, 1).
2. Possible arrangements with a sum of 6: (9, 9, 9), (9, 9, 6), (9, 6, 9), (6, 9, 9), (6, 6, 6).
So, there are six favorable outcomes.
Now, let's calculate the total number of possible outcomes. Since we are selecting three digits from four options without replacement, the number of possible outcomes is given by the combination formula:
Total outcomes = C(4, 3) = 4! / (3! * (4-3)!) = 4.
Therefore, there are four possible outcomes.
Finally, to find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes = 6 / 4 = 1.5.
However, it is important to note that probability must be a value between 0 and 1. Therefore, in this case, the probability would be 1 (or 100%) since it is not possible to have a fraction of an outcome.
So, the probability that the three randomly selected digits form a three-digit number that is a multiple of 3 is 1 (or 100%).