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Probability既難題 10分

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How many strings of 4 decimal digits (a) Do not contain the same digit twice? (b) End with an even digit? (c) Have 3 common digits? 麻煩列條式看看!!

最佳解答:

a) 9 * 9 * 8 * 7 = 4536 first digit not zero, second digit has 9 choices, may contain zero b) 9 * 10 * 10 * 10 / 2 = 4500 all numbers formed with repetition / 2, half even and half odd c) 10 * 9 * 4 - 9 - 9 * 3 = 324 3 digits same and the other not equal (but first digit may contain zero) = 10 * 9 * 4 numbers with first digit zero 0111, 0222 ...0999 ( 9 個 ) and 0100, 0010, 0001, 0200, 0020, 0002 ..... ( 9 * 3 個 )

其他解答:

4 decimal digits include the figures from .0000 to .9999; there are totally 10,000 figures. a. possibility of no same digit twice (from .0000 to .9999) = 10/10 * 9*10 * 8 /10 * 7/10 = 5,040 / 10,000 the number of strings of 4 decimal digits not contain the same digit twice = 5,039 b. there are 1/2 probability with an even digit from .0000 to .9999, that i.e 5,000 strings of decimal digits with an even digit. c. possibility of no same digit twice (from .0000 to .9999) = 10/10 * 1*10 * 1 /10 * 9/10 * 4 (“*4* means there are 4 cases with 3 common digit) = 360 / 10,000 The number of strings of decimal digits with 3 common digit = 3607638E748EED250B0