0
What else can I help you with?
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 choosing a vowel or one of the first 10 letters of the alphabet?
The probability for a single random choice, is 6/13.
What is the theoretical probability of landing 2 ones in a row with a 6 sided die?
In the first two throws, the probability is (1/6)*(1/6) = 1/36. Eventually landing them in repeated throws is 1 (a theoretical certainty).
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
How many 3-letter code words are possible using the first 6 letters of the alphabet if no letter can be repeated?
120
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
To deliver repeated blows plus 6 letters?
thrash
How many 3-letter codes are possible using the first 6 letters of the alphabet if letters can be repeated?
-3
What is the probability of choosing a vowel or one of the first 10 letters of the alphabet?
The probability for a single random choice, is 6/13.
What is the probability of writing correct addresses to 4 letters to be posted?
1/6
How many letters in the word letter?
Ther are 6 letters in the word letter two of which are repeated which gives a total of four different letters.
What is the theoretical probability of landing 2 ones in a row with a 6 sided die?
In the first two throws, the probability is (1/6)*(1/6) = 1/36. Eventually landing them in repeated throws is 1 (a theoretical certainty).
How many ways can you rearrange the letters in the word pencil?
To calculate the number of ways the letters in the word "pencil" can be rearranged, we first determine the total number of letters, which is 6. Since there are two repeated letters (the letter 'e'), we divide the total number of letters by the factorial of the number of times each repeated letter appears. This gives us 6! / 2! = 360 ways to rearrange the letters in the word "pencil."
6 In how many ways can the letters in the word SCRAMBLE be rearranged?
There are 8 letters with no repeated letters and you are working out the number of combintaions of all 8 letters with respect to order so the answer is 8P8=8!=40320
What A license plate consists of 2 letters followed by 4 digits letters and digits can be repeated. Find the probability that your new license plate begins with a vowel and ends with a multiple of 3.?
Independent events, so P(both)=(5/26)(4/10)=0.07692307692. The 4/10 comes from the fact that 0, 3, 6, and 9 are the 4 digits that are multiples of 3.
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
How many 3-letter code words are possible using the first 6 letters of the alphabet if no letter can be repeated?
120
