What is the probability that a number chosen at random from the range 1,000,000,000 to 9,999,999,999 inclusive will contain ten different digits? Round your answer to the nearest millionth
The number of ways the 10 distinct digits can form distinct numbers is given by
10!.
So the probability is 10! divided by the number of possible outcomes.
The first digit on the right can be any number from 0 to 9 so ten choices
1/10 so far
the second digit (tens place) can be one of 9
1/10 * 1/9
the third can be one of 8 remaining
1/10 * 1/9 * 1/8
etc
get the picture?
So after i do all that should i get one whole number
No, you will not get a whole number
a probability is a number between 0 and 1
1/10! which is about .00000027
To find the probability that a number chosen at random from the given range will contain ten different digits, we need to determine the number of possibilities for such numbers and divide it by the total number of possible numbers in the range.
First, let's determine the number of possibilities for a number containing ten different digits. We have ten slots to fill with ten different digits, so we can choose a digit for the first slot in 10 ways, the second slot in 9 ways (since it should be different from the first digit), the third slot in 8 ways (different from the first two digits), and so on until the tenth slot, which has only one possibility. This can be expressed as:
10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 10!
Next, let's determine the total number of possible numbers in the given range. The range spans from 1,000,000,000 to 9,999,999,999, inclusive. So, the total number of possible numbers is the difference between the highest and lowest numbers, plus one:
9,999,999,999 - 1,000,000,000 + 1 = 9,000,000,000
Now, to find the probability, we divide the number of possibilities for numbers with ten different digits by the total number of possible numbers:
P = (10! / 9,000,000,000)
Therefore, the probability is approximately 0.000000333, rounded to the nearest millionth.