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A license plate number is created by writing a sequence of three lette

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A license plate number is created by writing a sequence of three lette  [#permalink]

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New post 15 Jun 2018, 22:36
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A license plate number is created by writing a sequence of three letters followed by a sequence of 3 digits. Any digit 0 - 9 is acceptable, as is any letter. What is the probability that either (but not both) the letters or numbers form a palindrome, a series of letters or numbers that read the same forwards and backwards? (For example, ADA and 121 are palindromes but ABC and 123 are not. ADA121 would not be valid but ADA123 would be valid.)

A. 1/26
B. 1/10
C. 17/130
D. 7/52
E. 9/65

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A license plate number is created by writing a sequence of three lette  [#permalink]

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New post 16 Jun 2018, 02:01
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Alphabet palindrome * no. not a palindrome + Alphabet not a palindrome * no. a palindrome.

For alphabet to be a Palindrome, lets take one alphabet for ex: A,any of the 26 letters,A. For 26 alphabets it should be 26(1*26*1) = 26^2

For alphabet not to be a palindrome, lets take one alphabet for ex: A, any of the 26 letters, any of the 25 letters except A.
For 26 Alphabets it should be 26(1*26*25)

For number to be a palindrome, same type of calculation will result in: 10(1*10*1)
For number to be not a palindrome: 10(1*10*9)

total = 26(1*26*1)*10(1*10*9) + 26(1*26*25)*10(1*10*1) = 26^2 * 10^2*9 + 26^2 * 25 * 10^2

Total possible combinations = 26^3 * 10^3

Probability = Total / Total possible combinations

= (26^2 * 10^2*9 + 26^2 * 25 * 10^2)/26^3 * 10^3

= 17/130

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Re: A license plate number is created by writing a sequence of three lette  [#permalink]

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New post 29 Jun 2018, 11:55
This would be the probability of having a letter palindrome P(LP) OR the probability of having a number palindrome P(NP) less the probability of having a letter palindrome P(LP) AND the probability of having a number palindrome P(NP):

P(LP)= 26*(1/26 * 26/26 *1/26)= 1/26
P(NP)= 10*(1/10*10/10*1/10)=1/10

P(LP u NP)= 1/26 +1/10 - (1/26*1/10)= 35/260

Now, we still need to substract P(LP n NP), because in P(LP u NP) we are counting the outcome in which both happen at the same time and we are required to calculate either one or the other BUT NOT BOTH, so...

P(LP u NP) - P(LP n NP)= 35/260 - 1/260 = 17/130
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Re: A license plate number is created by writing a sequence of three lette  [#permalink]

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New post 29 Jun 2018, 19:48
Bunuel wrote:
A license plate number is created by writing a sequence of three letters followed by a sequence of 3 digits. Any digit 0 - 9 is acceptable, as is any letter. What is the probability that either (but not both) the letters or numbers form a palindrome, a series of letters or numbers that read the same forwards and backwards? (For example, ADA and 121 are palindromes but ABC and 123 are not. ADA121 would not be valid but ADA123 would be valid.)

A. 1/26
B. 1/10
C. 17/130
D. 7/52
E. 9/65


a quick answer..
Since the palindromes are only 3 letters or digits, the probability would depend on the third digit/letter

so

1) \(L_1-L_2-L_3-D_1-D_2-D_3\)---------- 26-26-1-10-10-9
when palindrome is with letters, the 3-digit number can have any of remaining 9 digits in units digit, as it has to be different from the first digit \(D_1\)
2) \(L_1-L_2-L_3-D_1-D_2-D_3\)---------- 26-26-25-10-10-1
when palindrome is with digits, the 3-letter formed can have 25 ways in third place, \(L_3\), different from the first digit \(L_1\)
so total ways 9+25=34

while in total ways these number will be 26*10=260

P = \(\frac{34}{260}=\frac{17}{130}\)

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