Probability of getting a repeated digit = no. of favorable outcomes/total no. of possible outcomes
thus no. of favorable outcomes = 10
Considering that anyone of the 10 digits may appear as the first numeral as well as d last numeral,
No. of possible outcomes=10*10=100
hence probability of a repeated digit=10/100=0.1
Probability of getting a repeated letter = no. of favorable 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 different. So the next four blanks can be filled in 25*24*23*22ways.
no. of favorable outcomes = 26*25*24*23*22
Probability of getting a repeated letter=(26*25*24*23*22)/(26^6)
Total probability of getting A repeated digit OR letter=0.1+0.0255=0.1255
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