posted by Mary .
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.