Inclusion-Exclusion

Question

$N$ people toss their hat into a bin, randomly shuffled, returned one hat to each person. What is the probability that $5$th person got his own hat?

Comments

@srestha ma'am,  if 5th person got his own hat,then it doesn't necessarily mean that others didn't get their hat; they may get their own hat or may not get it..nothing is mentioned in the question about the other members.

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@Somoshree Datta 5

yes, it can be do with that approach also.

 

$ {\color{Red} {\frac{(n-1)}{n}\;*\;\frac{(n-2)}{(n-1)}\;*\;\frac{(n-3)}{(n-2)}\;*\;\frac{(n-4)}{(n-3)}\;*\;}} {\color{Green} {\frac{(1)}{(n-4)}\;*\;}} {\color{Blue} {\frac{(n-5)}{(n-5)}\;*\;\frac{(n-6)}{(n-6)}\;*\;\frac{(n-7)}{(n-7)}\;*\;.....\frac{(1)}{(1)}\;}} = \frac{1}{n}$

 

@srestha mam

when u r telling 5th person got his own hat, that means u also have to determine all other not getting their own hat.

How r u determining this??

 i just fixed only, 5th person... i didn't restrict any other person.

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$\Large \frac{\text{number of onto functions with cardinality n-1}}{\text{number of onto functions with cardinality n}}  = \frac{(n-1)!}{n!} = \frac{1}{n}$

 

EDIT:-

Note that, no.of elements( let assume p ) are equal in A and B, then no.of onto functions are p! and every onto function is one-to-one also

( think why this is true?)