Probability theory question: randomly select integers from 1 to 2000. What's the probability that the integers can't be divided by 6 or 8? The answer is 1917 / 2000

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• 1. Take a random number from the integers of 1-2000. What's the probability that the integers can't be divided by 6 or 8 Because 333 is less than 2000 / 6 and less than 334, P (a) = 333 / 2000! Why should p (a) be expressed in this way and 333 / 2000?
• 2. Take a number randomly from the integers of 1 &; 2000. What's the probability that the number can't be divided by 6 or 8?
• 3. The probability of being divisible by 2 or 3 in 10-99,
• 4. Take any three of the 10 numbers from 0 to 9 to form a three digit number without repetition. The probability that this number cannot be divided by 3 is______ ．
• 5. Take any number from 1 to 100 and find the probability that the number can be divided by 2 or 3
• 6. Take any one of the 100 integers from 1 to 100, and try to find the probability that the number can be divided by 6 or 8
• 7. Take any one of 100 integers and calculate the probability that this number can not be divided by 6 or 8
• 8. Write the numbers 1, 2, 3, 4 and 5 on the five cards, mix them and arrange them in a row. What is the probability that the number can be divided by 2 or 5? Why is the last place divisible by 2 only 2 or 4? There are 48 kinds of 2 * 4 The last place divisible by 5 can only be 5, total = 24 kinds How did you get here?
• 9. 5_ 383_ 8_ 2_ 936_ 5_ 8_ 203_ 9_ 3_ The reason why the probability of dividing 396 by any number 0 ~ 9 added at the line is 1
• 10. The probability that any number from 1 to 100 can be divided by 5 or 7
• 11. In the 1 to 2000 random integer, ask the probability that the integer can not be divided by 6 or 8
• 12. The probability that the product of the three integers can be divisible by 10 is obtained by randomly taking three times from nine integers from one to nine, one at a time I read the answers on the Internet, but I can't understand them, especially the ones with one five and two five. I don't know how to calculate the probability I thought there must be a 5 that is 1 / 9, there must be an even number 4 / 9, another random 9 / 9, but the multiplication is not right 214, why not? It's best to write the reason step Otherwise I can't understand it
• 13. From 1 to 9, there are nine numbers that are put back. Take three times, one at a time. The product of three numbers can be divided by 10
• 14. Take any one from 1,2,3,... To 9, record and put it back, take n times continuously, and find the probability that the product of the number taken can be divided by 10?
• 15. If six digit 2003 () () can be divided by 45, what are its last two digits? If four digit 36ab can be divided by 2,3,4,5,9 at the same time, what is 36ab? Add 10 points to answer the second question
• 16. If the six digit 1992 can be divided by 105, then the last two digits are______ ．
• 17. Six digit 2003 can be divided by 99, and its last two digits are______ ．
• 18. Six digit 2003 can be divided by 99, and its last two digits are______ ．
• 19. Six digit 2006 can be divided by 99, and its last two digits are_____________ Process + answer
• 20. Write a positive integer whose sum of all the numbers in the proposition is a multiple of 3, an inverse proposition, a no proposition, an inverse no proposition that can be divided by 3, and judge whether it is true or false