If a digit can be used more than once, what is the
probability that the last two digits of a randomly generated 5-digit
number are both 4?
A. 1/100 B. 1/90 c. 1/81 D. 1/10
What is your answer?
so, in the numbers from 00-99
how many are 44?
1? huh
right. So, 1/100 chance of getting it.
To find the probability that the last two digits of a randomly generated 5-digit number are both 4, we need to consider the total number of possible 5-digit numbers and the number of those numbers where the last two digits are both 4.
Let's break down the problem step-by-step:
Step 1: Determine the total number of possible 5-digit numbers.
Since each digit can be used more than once, any digit from 0 to 9 can be used in each of the five positions. Therefore, there are 10 possible choices for each of the five positions, resulting in a total of 10^5 = 100,000 possible 5-digit numbers.
Step 2: Determine the number of 5-digit numbers where the last two digits are both 4.
In this scenario, we want the last two digits to be both 4. This means that the other three digits can be any digit from 0 to 9, but they are not restricted to being 4. So we have 10 choices for each of the first three positions, and only one choice for each of the last two positions (which must be 4). Therefore, there are 10^3 * 1 * 1 = 10^3 = 1,000 5-digit numbers where the last two digits are both 4.
Step 3: Calculate the probability.
The probability is determined by dividing the number of successful outcomes (the number of 5-digit numbers where the last two digits are both 4) by the total number of possible outcomes (the total number of 5-digit numbers).
Therefore, the probability is: 1,000 (successful outcomes) / 100,000 (total outcomes) = 1/100
So the correct answer is A. 1/100.