Assume that 3 digits are selected at random from the set {1,3,5,6,7,9} and are arranged in random order.

What is the probability that the resulting 3-digit number is less than 700?

I know the sample space is P(6,3) but that is all I know. Help!

You are correct that the number of possible numbers is the number of permutations, which is 6!/3! = 120

Of those numbers, 1/3 will start with 7 or 9. Those numbers will exceed 700 and the others will not. So the answer is 40.

To calculate the probability that the resulting 3-digit number is less than 700, we first need to determine the total number of possible outcomes, and then determine the number of favorable outcomes.

Total number of possible outcomes:
The set {1, 3, 5, 6, 7, 9} contains 6 digits. We need to select 3 digits from this set to form a 3-digit number. Since repetition is not allowed, we can use the concept of permutations. The formula for permutations of k items from n items is given by P(n, k) = n! / (n-k)!. In this case, we have n = 6 (number of digits in the set) and k = 3 (number of digits needed to form a 3-digit number). Therefore, the total number of possible outcomes is P(6, 3) = 6! / (6-3)! = 6! / 3! = 120 / 6 = 20.

Number of favorable outcomes:
To form a 3-digit number less than 700, the hundreds digit must be either 1, 3, 5, or 6. We need to consider two cases:

Case 1: When the hundreds digit is 1.
In this case, there are 5 remaining digits to choose from for the tens and units place (3, 5, 6, 7, 9). By the same permutation argument, the number of favorable outcomes is P(5, 2) = 5! / (5-2)! = 5! / 3! = 20 / 2 = 10.

Case 2: When the hundreds digit is 3, 5, or 6.
In this case, there are 4 remaining digits to choose from for the tens and units place (1, 5, 7, 9). Similarly, the number of favorable outcomes is P(4, 2) = 4! / (4-2)! = 4! / 2! = 12 / 2 = 6.

Adding the favorable outcomes from both cases, we get a total of 10 + 6 = 16 favorable outcomes.

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes = 16 / 20 = 0.8 or 80%.

Therefore, the probability that the resulting 3-digit number is less than 700 is 0.8 or 80%.

To find the probability that the resulting 3-digit number is less than 700, we need to determine two things:

1. The total number of possible outcomes (sample space).
2. The number of favorable outcomes (the numbers that meet the given condition).

Let's start with the sample space.

Given that we have 6 digits to choose from (1, 3, 5, 6, 7, and 9), and we need to select 3 digits to form a 3-digit number, we can use the permutation formula.

The number of ways to select and arrange 3 digits out of 6 is calculated as P(6, 3) = 6! / (6 - 3)! = 6! / 3! = 6 * 5 * 4 = 120.

So, the sample space consists of a total of 120 possible outcomes.

Next, let's determine the number of favorable outcomes (the numbers less than 700).

To create a number less than 700, the hundreds digit must be either 1, 3, 5, or 6. The tens and units digits can be any of the available digits (1, 3, 5, 6, 7, and 9).

For the hundreds digit, we have 4 options (1, 3, 5, and 6).
For the remaining tens and units digits, we have 5 options (since we cannot repeat the hundreds digit in the other two positions).

Therefore, the number of favorable outcomes is 4 * 5 * 5 = 100.

Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes
= 100 / 120
= 5 / 6

So, the probability that the resulting 3-digit number is less than 700 is 5/6, approximately 0.8333, or 83.33%.