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What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has one numeral follwed by seven letters followed by one numeral?
Wiki User
∙ 11y ago
This is easiest to answer by working out the probability of not getting such a license plate and subtracting that from 1.
There are 10 x 26 x 26 x 26 x 26 x 26 x 26 x 26 x 10 different possible license plates (since there are 10 digits: 0-9 and 26 letters: A-Z).
To not have any repeated digits or letters, there are: 10 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 9 possible license plates (since the two digits must be different and the seven letters must all be different).
Meaning the probability of getting one of these is:
(10 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 9)/(10 x 26 x 26 x 26 x 26 x 26 x 26 x 26 x 10) = (25 x 3 x 23 x 11 x 21 x 9)/(26 x 13 x 13 x 13 x 13 x 13)
= 3586275/9653618
⇒ probability of getting at least one repeating digit or number is:
1 - 3586275/9653618 = 6067343/9653618≈ 0.629
Wiki User
∙ 11y ago
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What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has three numerals followed by two letters followed by three numerals?
This is one minus the probability of having no repeated letters. The probability of having no repeated numbers is 10/10 x 9/10 x 8/10 x 7/10 x 6/10 x 5/10 which equals 0.1512. The probability of having no repeated letters is 26/26 x 25/26 = 0.96153846153846153846153846153846 These multiplied together gives 0.14538461538461538461538461538462 1 minus this is 0.85461538461538461538461538461538 And so the probability of having a repeater letter or digit is roughly 0.855
What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has one numeral followed by six letters followed by one numeral?
Probability of getting a repeated digit = no. of favourable outcomes/total no. of possible outcomes Favourable outcomes=(0,0),(1,1),(2,2).....(9,9) thus no. of favourable outcomes = 10 Considering that anyone of the 10 digits may apperar as the first numeral as well as d last numeral, No. of possible outcomes=10*10=100 hence probablity of a repeated digit=10/100=0.1 Probability of getting a repeated letter = no. of favourable outcomes/total no. of possible outcomes consider 6 blanks _ _ _ _ _ _ ,each filled with a letter. Thus no. of possible outcomes = 26^6 now consider that any two of these blanks have the same letter. Consider the two blanks filled with same letter as one blank. _ _ _ _ _ So that blank can be filled in 26 ways(i.e. you can have any of 26 alphabets as the repeated letter) The other 4 blanks can be filled with rest of 25 alphabets as the one that has already been used(the repeated letter) cannot be used again. We want all other letters to be differrent. So the next four blanks can be filled in 25*24*23*22ways. Thus, no. of favaourable outcomes = 26*25*24*23*22 Probability of getting a repeated letter=(26*25*24*23*22)/(26^6) =0.0255(approx) And, Total probablity of getting A repeated digit OR letter=0.1+0.0255=0.1255 i.e. 12.55% __________ Another approach, if the letters are able to be recurring throughout, which they usually are, you have a 1 in 95,428,956,661,682,176 of getting two of the same letters right next to each other. You first take 26^6, then it comes out as a 1 in 308,915,776 of getting a letter you want. Then you square that, because the outcome MUST be the same, so the chances of you getting that same letter increase by whatever the denominator is, which would be 308,915,776^2. Hopefully I'm right. xD
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What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has three numerals followed by two letters followed by three numerals?
This is one minus the probability of having no repeated letters. The probability of having no repeated numbers is 10/10 x 9/10 x 8/10 x 7/10 x 6/10 x 5/10 which equals 0.1512. The probability of having no repeated letters is 26/26 x 25/26 = 0.96153846153846153846153846153846 These multiplied together gives 0.14538461538461538461538461538462 1 minus this is 0.85461538461538461538461538461538 And so the probability of having a repeater letter or digit is roughly 0.855
What is the probability of getting a license plate with a repeated letter or digit if you live in a state that has a 3 letters followed by 2 numbers?
The number of license plates without a repeated letter or digit is 26x25x24x10x9 = 1 404 000. The number of license plates without any restriction is 26x26x26x10x10 = 1 757 600. Probability of a license plate without a repeated letter or digit = 1 404 000 /1 757 600 = 0.7988 Probability of a license plate with a repeated letter or digit = 1-0.7988 = 0.2022
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What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has one numeral followed by six letters followed by one numeral?
Probability of getting a repeated digit = no. of favourable outcomes/total no. of possible outcomes Favourable outcomes=(0,0),(1,1),(2,2).....(9,9) thus no. of favourable outcomes = 10 Considering that anyone of the 10 digits may apperar as the first numeral as well as d last numeral, No. of possible outcomes=10*10=100 hence probablity of a repeated digit=10/100=0.1 Probability of getting a repeated letter = no. of favourable outcomes/total no. of possible outcomes consider 6 blanks _ _ _ _ _ _ ,each filled with a letter. Thus no. of possible outcomes = 26^6 now consider that any two of these blanks have the same letter. Consider the two blanks filled with same letter as one blank. _ _ _ _ _ So that blank can be filled in 26 ways(i.e. you can have any of 26 alphabets as the repeated letter) The other 4 blanks can be filled with rest of 25 alphabets as the one that has already been used(the repeated letter) cannot be used again. We want all other letters to be differrent. So the next four blanks can be filled in 25*24*23*22ways. Thus, no. of favaourable outcomes = 26*25*24*23*22 Probability of getting a repeated letter=(26*25*24*23*22)/(26^6) =0.0255(approx) And, Total probablity of getting A repeated digit OR letter=0.1+0.0255=0.1255 i.e. 12.55% __________ Another approach, if the letters are able to be recurring throughout, which they usually are, you have a 1 in 95,428,956,661,682,176 of getting two of the same letters right next to each other. You first take 26^6, then it comes out as a 1 in 308,915,776 of getting a letter you want. Then you square that, because the outcome MUST be the same, so the chances of you getting that same letter increase by whatever the denominator is, which would be 308,915,776^2. Hopefully I'm right. xD
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