A 3-digit positive integer is selected at random. Find the probability that
A) The last two digits of the number is 22
B) The number is greater than 900
C) The number is less than 100
D) The number contains at least one digit 8
there 900 3-digit integers
(a) only 9 end in 22: 122,222,...,922
Try the others
A) The last two digits of the number is 22.
Well, 22 is a pretty epic number, but how many 3-digit numbers end with 22? Let's see... There are 10 choices for the first digit (1 through 9, excluding 0) and 1 choice for the second digit (2). So, we have 10 * 1 = 10 possibilities that end with 22.
But how many total 3-digit numbers are there? Well, remember that the first digit cannot be 0, so we have 9 choices for the first digit (1 through 9), each digit has 10 possibilities, so the total number of 3-digit numbers is 9 * 10 * 10 = 900.
Now we can find the probability by dividing the number of successful outcomes (10) by the total number of possibilities (900):
Probability = 10/900 = 1/90
So, the probability that the last two digits of the number is 22 is 1/90.
B) The number is greater than 900.
To find the probability, we need to count the number of 3-digit numbers that are greater than 900.
Since we need the first digit to be 9 or greater, there are only two choices: 9 or 10. For the second and third digits, each digit has 10 possibilities. So, we have 2 * 10 * 10 = 200 possibilities.
Using the total number of 3-digit numbers (900) as the denominator, we have:
Probability = 200/900 = 20/90 = 2/9
Therefore, the probability that the number is greater than 900 is 2/9.
C) The number is less than 100.
Since the number needs to be less than 100, the first digit must be 0, and we have 1 choice for the first digit (0) and 10 choices for the second and third digits (0 through 9). So, we have 1 * 10 * 10 = 100 possibilities.
Again, using the total number of 3-digit numbers (900) as the denominator, we have:
Probability = 100/900 = 1/9
Therefore, the probability that the number is less than 100 is 1/9.
D) The number contains at least one digit 8.
To find the probability, we need to count the number of 3-digit numbers that contain at least one digit 8.
Let's think about this creatively. There are three positions for digits in a 3-digit number: the hundreds place, the tens place, and the units place. For each position, we have 10 choices (0 through 9). So, the total number of 3-digit numbers is 10 * 10 * 10 = 1000.
Now let's think about how many numbers don't contain digit 8. There are 9 choices for each position (0 through 7 or 9), so the total number of 3-digit numbers without digit 8 is 9 * 9 * 9 = 729.
Therefore, the number of 3-digit numbers that contain at least one digit 8 is 1000 - 729 = 271.
Using the total number of 3-digit numbers (1000) as the denominator, we have:
Probability = 271/1000 = 0.271
Therefore, the probability that the number contains at least one digit 8 is approximately 0.271.
To find the probability of each event, we first need to determine the total number of possible outcomes.
In this case, we are dealing with 3-digit positive integers, so the total number of possible outcomes is 900 (ranging from 100 to 999).
A) The last two digits of the number is 22.
To find the number of outcomes that satisfy this condition, we need to count how many 3-digit positive integers have 22 as the last two digits. This would be any number in the form of X22, where X can be any digit from 1 to 9 (excluding 0). There are 9 such possibilities (122, 222, 322, ..., 922).
So the probability of the last two digits being 22 is 9/900, which simplifies to 1/100.
B) The number is greater than 900.
To find the number of outcomes that satisfy this condition, we need to count how many 3-digit positive integers are greater than 900. This would be any number in the form of 9XY, where X and Y can be any digit from 0 to 9. There are 10 possibilities for X (0-9), and 10 possibilities for Y (0-9) since we can repeat digits.
So the probability of the number being greater than 900 is 100/900, which simplifies to 1/9.
C) The number is less than 100.
To find the number of outcomes that satisfy this condition, we need to count how many 3-digit positive integers are less than 100. Since we are dealing with positive integers, there are no 3-digit positive integers less than 100.
So the probability of the number being less than 100 is 0/900, which is 0.
D) The number contains at least one digit 8.
To find the number of outcomes that satisfy this condition, we need to count how many 3-digit positive integers contain at least one digit 8. There are two ways to approach this:
1) Count the number of outcomes that do not contain any 8's.
To do this, we need to count how many 3-digit positive integers have no 8's as any of their digits. Since each digit can be any digit from 1 to 9 (excluding 0), there are 9 choices for each of the three digits. Therefore, the number of outcomes without any 8's is 9^3 = 729.
2) Subtract the number of outcomes without any 8's from the total number of outcomes.
The total number of outcomes is 900 (as mentioned earlier). Subtracting the outcomes without any 8's (729) gives us the number of outcomes with at least one digit 8: 900 - 729 = 171.
So the probability of the number containing at least one digit 8 is 171/900, which simplifies to 19/100.
To recap:
A) The probability that the last two digits of the number are 22 is 1/100.
B) The probability that the number is greater than 900 is 1/9.
C) The probability that the number is less than 100 is 0.
D) The probability that the number contains at least one digit 8 is 19/100.
To find the probability for each scenario, we need to determine the total number of possible outcomes and the number of favorable outcomes for each case.
A) The last two digits of the number is 22:
There are 10 possible options for the first digit (0-9), and only one option for the third digit (2). So, the favorable outcomes are when the randomly selected number has the last two digits as 22. Therefore, the probability is:
Favorable Outcomes: 1
Total Possible Outcomes: 10 * 10 * 1 (three digits, each ranging from 0 to 9)
Probability: Favorable Outcomes / Total Possible Outcomes
B) The number is greater than 900:
Since we have a three-digit positive integer, the first digit must be 9. The second and third digit can be any number from 0 to 9. So, the favorable outcomes are when the randomly selected number is greater than 900. The probability is:
Favorable Outcomes: 1 * 10 * 10 (first digit is 9, second digit can be 0-9, and third digit can be 0-9)
Total Possible Outcomes: 10 * 10 * 10 (three digits, each ranging from 0 to 9)
Probability: Favorable Outcomes / Total Possible Outcomes
C) The number is less than 100:
For a three-digit positive integer, if the number is less than 100, the first digit must be 0. The second and third digit can be any number from 0 to 9. So, the favorable outcomes are when the randomly selected number is less than 100. The probability is:
Favorable Outcomes: 1 * 10 * 10 (first digit is 0, second digit can be 0-9, and third digit can be 0-9)
Total Possible Outcomes: 10 * 10 * 10 (three digits, each ranging from 0 to 9)
Probability: Favorable Outcomes / Total Possible Outcomes
D) The number contains at least one digit 8:
To count favorable outcomes, let's count the total outcomes where the number does not contain any digit 8 (complementary counting method), and then calculate the number of outcomes that do not meet this condition. The probability is:
Favorable Outcomes: Total Possible Outcomes - Outcomes without any digit 8
Total Possible Outcomes: 10 * 10 * 10 (three digits, each ranging from 0 to 9)
Probability: Favorable Outcomes / Total Possible Outcomes
To find the number of outcomes without any digit 8, we can calculate the number of outcomes where each digit is not 8. For each digit, there are 9 options (0-9 excluding 8). The probability is:
Outcomes without any digit 8: 9 * 9 * 9 (each digit can take on any value from 0 to 9 besides 8)
Now, we can plug these values into the respective formulas to get the probabilities for each scenario.