Probability Question 66
Question: A mapping is selected at random from the set of all the mappings of the set $ A={ 1,2,…,n } $ into itself. The probability that the mapping selected is an injection is
Options:
A) $ \frac{1}{n^{n}} $
B) $ \frac{1}{n!} $
C) $ \frac{(n-1)!}{{n^{n-1}}} $
D) $ \frac{n!}{{n^{n-1}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
The total number of functions from $ A $ to itself is $ n^{n} $ and the total number of bijections from $ A $ to itself is $ n!. $ {Since $ A $ is a finite set, therefore every injective map from $ A $ to itself is bijective also}.
$ \therefore $ The required probability $ =\frac{n!}{n^{n}}=\frac{(n-1)!}{{n^{n-1}}}. $
BETA
