1i) \frac{(n-1)!}{(n-1)^n} ?
consider a town with n people.
a person spreads a rumour to a second,who in turn repeats it to a third and so on.suppose that at each stage,the recipient of the rumour is chosen at random from the remaining (n-1) people.
what is the probability that the rumour will be repeated n times (i)without being repeated to any person
(ii)without being repeated to the originator
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12 Answers
39ii)
the originator has (n-1) chioces and all (n-1) others have (n-2) choices.
so no:of ways it can be done is
(n-1)(n-2)(n-1)
total no: of ways =(n-1)n
so probability = [(n-2)/(n-1)](n-1)
1thank u b555....
wow its the first time someone replied correctly to a question at
01:02 am....
:-)
106i think (i) is zero because for n repetitions of the rumour, there need to be n+1 people without any people getting repeated.
So there is no case that the rumour will be repeated n times without being repeated to any person
Consider a simple ex.
there are 5 people
1-->2 --> 3-->4 -->5 so without repetition of people there are 4 repetitions of the rumour i.e. (n-1)
So the prob. is zero
if u count 5-->1 as a repetition then look below
1nishant sir,kaymant sir or prophet sir can u pls justify asish's answer because i dont have the answer to this question
pls
106oh.. yes i assumed that since 1 knows it, it will be considered as a repetition.
In that case, it will be (n-1)(n-1)(n-3)!(n-1)n
this is bcz, the first person has (n-1) ways so has the second person but then the third has (n-3) ways and this keeps on decreasing by one
62I go with the first answer of asish.. which is zero..
To be mreo clear we need the question to mention what "repeat" means.. To me what asish has doen there makes a lot of sense..