Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Assume that numbers can start with 0. What is the probability of choosing a random number that starts with 8 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary.

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gwibdaithq · Accepted Answer

It is best if the first and last digits match. Therefore, the total number of potential numbers is 5*4*8*7 = 1120.Total amount that begins with2=4⋅8⋅7=224Hence, P(Stars with 2)=2241120=15

Nick Camelot · Accepted Answer

Step 1:First, let&#039;s consider the possibilities for the first digit. Since it must be even, we have two choices: 0 or 8.For the last digit, we have five choices since it can be any even digit other than the one already chosen for the first digit.Now, let&#039;s look at the choices for the remaining two digits in the middle. We have eight digits left (1, 2, 3, 4, 5, 6, 7, 9) since we have already used two even digits. For the first remaining digit, we have eight choices, and for the second remaining digit, we have seven choices.By the counting principle, we multiply the number of choices for each digit together: 2×5×8×7=560.Step 2:Next, we need to determine the total number of four-digit numbers that can be created using the digits 0-9, without any restrictions. This can be found by multiplying the number of choices for each digit: 10×9×8×7=5,040.Finally, we can calculate the probability of choosing a random number that starts with 8 by dividing the number of favorable outcomes (numbers that start with 8) by the total number of possible outcomes:5605,040=0.1111Rounded to four decimal places, the probability is 0.1111.

Mr Solver · Accepted Answer

The first digit can be chosen from the set {0,2,4,6,8} (5 options), as it must be even. The last digit can also be chosen from the same set (5 options). The remaining two digits can be chosen from the set {0,1,2,3,4,5,6,7,8,9}, excluding the digits already chosen for the first and last positions (8 options for the first remaining digit, and 7 options for the second remaining digit).Therefore, the total number of possible four-digit numbers meeting the given criteria is 5×5×8×7=1400.Now, let&#039;s determine the number of four-digit numbers that start with 8 and meet the given criteria. The first digit is fixed to be 8, and the last digit can be chosen from the set {0,2,4,6,8} (5 options). The remaining two digits can be chosen from the set {0,1,2,3,4,5,6,7,9} (9 options for the first remaining digit, and 8 options for the second remaining digit).Therefore, the number of four-digit numbers that start with 8 and meet the criteria is 1×5×9×8=360.Finally, we can calculate the probability of choosing a random number that starts with 8 from this group by dividing the number of favorable outcomes (numbers that start with 8) by the total number of possible outcomes.Thus, the probability is 3601400=0.2571 (rounded to 4 decimal places).Hence, the probability of choosing a random number that starts with 8 from this group is 3601400=935.

madeleinejames20 · Accepted Answer

Result:15Solution:Let&#039;s denote the four-digit numbers as abcd, where a, b, c, and d represent the digits. We have the following conditions:- The first and last digits must be even: a and d are even digits. There are 5 even digits to choose from (0, 2, 4, 6, 8).- No digit can repeat: a, b, c, and d are distinct digits. We have 10 digits in total (0-9), and since a and d are already determined to be even digits, we have 8 remaining digits to choose from for b and c.Therefore, the total number of four-digit numbers that satisfy the conditions is 5×8×8×1.Now, let&#039;s consider the number of four-digit numbers that start with 8. In this case, a is fixed to be 8, and the remaining digits b, c, and d can be chosen from the 8 remaining digits (0, 1, 2, 3, 4, 5, 6, 9). Therefore, the number of four-digit numbers that start with 8 is 1×8×8×1.Finally, we can calculate the probability by dividing the number of four-digit numbers that start with 8 by the total number of four-digit numbers:P({number&amp;nbsp;starts&amp;nbsp;with&amp;nbsp;8})=1×8×8×15×8×8×1=15=15.Hence, the probability of choosing a random number that starts with 8 from this group is 15.

Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat.

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