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consider all four digit numbers that can be made from the digits 0-9 ( assume that numbers cannot start with zero). what is the probability of choosing a random number from this group that is less than or equal to 6000? Enter a fraction or round your answer to 4 decimal places.

User Kritzefitz
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1 Answer

5 votes

Answer:

Prob. of < 6000: 2/3

Explanation:

To find the probability of choosing a random number from the group of all four-digit numbers (without leading zeros) that is less than or equal to 6000, we need to determine the total number of valid numbers and divide it by the total number of possible numbers.

First, let's calculate the total number of possible numbers without leading zeros. Since there are 10 digits (0-9) available for each position, there are 9 choices for the thousands place (1-9) and 10 choices for each of the other three positions (0-9). Thus, the total number of possible numbers is 9 * 10 * 10 * 10 = 9,000.

Now, let's determine the total number of valid numbers less than or equal to 6000. Since the leading digit must be 5 or less, there are 6 choices for the thousands place (1-6). For the other three positions, each digit can be any number from 0 to 9. Thus, the total number of valid numbers is 6 * 10 * 10 * 10 = 6,000.

To calculate the probability, we divide the number of valid numbers by the total number of possible numbers:

Probability = Number of valid numbers / Total number of possible numbers

Probability = 6,000 / 9,000

Probability = 2/3 or 0.6667 (rounded to 4 decimal places)

Therefore, the probability of choosing a random number from this group that is less than or equal to 6000 is 2/3 or approximately 0.6667.

User Paul Mooney
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