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## Main Question or Discussion Point

A code is generated consisting of 3 letters followed by 2 digits. Each of the 3 letters generated is equally likely to be any of the 26 letters of the alphabet. Each of the 2 digits generated is equally likely to be any of the nine digits 1-9. Find the probability that a randomly chosen code has exactly 2 letters the same or 2 digits the same, but not both.

Attempted solution:

P(exactly 2 digits same) = (26

P(exactly 2 letters same) = (26*25*3*9

P(2 digits same AND 2 letters same) = (26*25*3*9)/(26

Required probability = 1/9 + 75/676 - 25/2028 = 0.210

But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...

Thanks!

Attempted solution:

P(exactly 2 digits same) = (26

^{3}*9)/(26^{3}*9^{2}) = 1/9P(exactly 2 letters same) = (26*25*3*9

^{2})/(26^{3}*9^{2}) = 75/676P(2 digits same AND 2 letters same) = (26*25*3*9)/(26

^{3}*9^{2}) = 25/2028Required probability = 1/9 + 75/676 - 25/2028 = 0.210

But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...

Thanks!