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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

Q: 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?

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The question is ambiguous. It is in complete.

The probability is 1. The letters in the word mathematics are all constants, not variables!

It is (26/26)*(25/26)*(24/26)*(23/26) = 358800/459676 = 0.79 approx.

Since the word "probability" contains only letters, then the probability of choosing a letter from the word "probability" is 1, i.e. it is certain to happen.

There are 9,979,200

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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

It is approx 54%.

3490 to 1

This may not be quite as straightforward as it first seems. In the UK, which has a different letter-number combination, some letters are not used so as to avoid confusion for example O and 0, or I and 1. If that is the case, then the probability of a particular digit is not 1/10. Also, some letter combinations that spell offensive words are disallowed. The consequence is that not all 26 letters of the alphabet are used and those that are may not be used in the same proportion. This will probably not apply if there are only two letters.

3/26

The question is ambiguous. It is in complete.

I don't know how to get past this point but the probability of having a letter appear more than once is: 98176/456796 = 21.4% The likelihood of a digit appearing more than once is: 280/1000=28% If they need to be consecutive to count then the letters would be: 50726/456796 = 11.1% The numbers would be: 100/1000 = 10% I don't know how to combine these probabilities properly so I'll leave that to you to figure out or for someone else to improve on.

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

The probability is 1. The letters in the word mathematics are all constants, not variables!

There are ten digits and twenty-six letters, assuming unconstrained choice of both. The probability of getting a repetition in the first two digits is 10/100, since there are 10 repetitious combinations of two digits among the 100 pairs of digits possible.To calculate the probability of a repetition in the two letters, consider that there are 262 combinations, of which 26 X 25 do not contain repetitions. (There is free choice among 26 alternatives for the first letter, but only 25 choice for the following letter.) Therefore, the probability of a repetition within the two letters is (26)(25)/262 or 25/26.The chance of a repeated digit within the four digit bloc is (10)(9)(8)(7)/104) or 504/1000.If there is no repetition of a digit within either the two-digit bloc or the four-digit bloc, there is still a chance of repetition of a digit between the blocs. The chance of this is (8 X 7 X 6 x 5)/104 or 168/1000.To reach the final answer, it is advantageous to calculate first the probability of not getting the specified repetition and then subtract this number from one. The probability of not getting a repetition in the first bloc is 9/10, and the probability of not getting a repetition between the first and second blocs is 168/1000. Therefore, the probability of not getting a repetition within the first bloc or between the first and the second number blocs is the product of these two, or 1512/10,000. Further multiplications by the independent probabilities of no repetition within the two remaining blocs yields a value of about 2.88 X 10-4. Therefore, the probability of at least one of the repetitions specified is about 0.999712.

It is (26/26)*(25/26)*(24/26)*(23/26) = 358800/459676 = 0.79 approx.

The letters that are repeated in the Spanish alphabet are "L", "A", "E", and "S."