Conditional Function Mapping

Probability Level pending

Consider the set of all functions from { 1 , 2 , 3 , 4 , 5 } \{1, 2, 3, 4, 5\} to itself that is not surjective (onto). Let f f be taken uniformly randomly from the set. Find the probability that there exists i { 1 , 2 , 3 , 4 , 5 } i \in \{1, 2, 3, 4, 5\} such that f ( i ) = i f(i) = i .

None of above 2069/3025 44/120 81/125 405/601

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1 solution

Prajwal Krishna
Nov 4, 2016

Total number of functions possible = 3125 As for each element in domain 5 choices are there, so total functions 5^5 Total number of onto function possible=120 As for first element there is 5 options , for second there is 4 options and so on. So total onto functions possible 5! hence total number of into functions 3125-120=3005

Now let us count total functions where f(i) \neq i For i=1 we have 4 options {2,3,4,5}, similarly we have 4 options for i=2{1,3,4,5} Hence total functions where f(i) \neq i=4^5=1024 However, out of these not all are into functions Let us count total onto functions where f(i) \neq i Since 1 should not be mapped with 1 , 2 should not be mapped with 2 and so on; further each of {1,2,3,45} should be uniquely mapped so number of onto functions where f(i) \neq i = Derangement of 5 =44 Hence, total onto functions where f(i) \neq i = 44 Therefore, total into functions where f(i) \neq i=1024-44 Hence, total into functions where f(i)=i for atleast one of i={1,2,3,4,5} =3005-(1024-44) = 2025 Hence required probability= 2025 3005 \frac { 2025 }{ 3005 } = 405 601 \frac { 405 }{ 601 }

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