a number between 1000 and 9999, inclusive, is chosen at random. what is the probability that it contains a)no 9's? b) at least one 9?
There are 9000 numbers from 1000 to 9999 inclusive
then number without any 9 are
8x9x9x9 = 5832
reasoning:
the lead digit has to be
1,2,3,4,5,6,7,8
after that each place can be filled in 9 ways, that is, with
0,1,2,3,4,5,6,7,8
prob (no 9 anywhere) = 5832/9000 = .648
so prob(at least one 9) = 1 - .648 = .352
To find the probability, we need to consider the total number of possible outcomes and the number of favorable outcomes for each case.
a) No 9's:
To find the number of four-digit numbers without any 9's, we can consider each digit separately.
The first digit can be any number from 1 to 8 since 9 is not allowed. Therefore, there are 8 options for the first digit.
The second, third, and fourth digits can each be any number from 0 to 8 since 9 is not allowed. So, there are 9 options for each of the three remaining digits.
To find the total number of four-digit numbers without any 9's, we multiply the number of options for each digit together:
8 * 9 * 9 * 9 = 5832
The total number of possible four-digit numbers is 9999 - 1000 + 1 = 9000.
Therefore, the probability of choosing a number between 1000 and 9999 (inclusive) with no 9's is: 5832/9000 = 0.648.
b) At least one 9:
To find the number of four-digit numbers with at least one 9, we can subtract the number of numbers with no 9's from the total number of four-digit numbers.
The total number of four-digit numbers is 9000.
From part a, we found that there are 5832 numbers with no 9's.
So, the number of four-digit numbers with at least one 9 is: 9000 - 5832 = 3168.
Therefore, the probability of choosing a number between 1000 and 9999 (inclusive) with at least one 9 is: 3168/9000 = 0.352.
To find the probability, we first need to determine the total number of four-digit numbers between 1000 and 9999, inclusive. Then we can count the number of numbers that meet the specific criteria for each part of the question.
a) No 9's:
We need to count the numbers with no 9's. In this case, we have 10 options (0-8) for each of the four digits (thousands, hundreds, tens, and units place). Thus, the total number of possibilities is 10*10*10*10 = 10,000.
b) At least one 9:
To determine the probability of having at least one 9, we will subtract the probability of having no 9's (found in part a) from 1.
Total number of possibilities (between 1000 and 9999) = 10,000.
So, the probability of having at least one 9 is 1 - (probability of no 9's) = 1 - (10,000/10,000) = 1 - 1 = 0.
Therefore, the probability of having at least one 9 is 0, meaning it is impossible to choose a number between 1000 and 9999 inclusive that does not have a 9.