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A.$\dfrac{1}{9}$

B.$\dfrac{1}{3}$

C.$\dfrac{1}{4}$

D.$\dfrac{2}{9}$

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If A and B are two non-empty sets, then a relation ‘f’ from set A to set B is said to be a function or mapping,

If every element of set A is associated with unique element of set B,

The mapping or function ‘f’ from A to B is denoted by f:A→A,

If f is a function from A to B and x ϵ A, then f(x) ϵ B where f(x) is called the image of x under f and x is called the preimage of f(x) under ‘f’.

Thus, for a mapping from A to B,

A and B should be non-empty.

Each element of A should have an image in B.

No element of ‘A’ should have more than one image in ‘B’.

Find the number of one-one maps and the total number of mappings from A to A.

$\therefore {\text{Required Probability = }}\dfrac{{{\text{Number of one - one map}}}}{{{\text{Total number of maps}}}}$

∴ A has 3-elements. Let it be a, b, c

∴ f :A→A

∴Number of one to one mapping $ = 3 \times 2 \times 1$

$ = 6$

And total number of mapping $ = 3 \times 3 \times 3$

$ = 27$

∴ Probability that the mapping is one to one is $ = \dfrac{{{\text{Number of one to one maps}}}}{{{\text{Total number of maps}}}}$

$ = \dfrac{2}{9}$

∴ Option (D) is correct

Every mapping is a relation but every relation may not be a mapping. If a set has n-elements and a mapping f is defined on A. Therefore, number of one-to-one mapping is \[\left| \!{\underline {\,

n \,}} \right. \]and the total number of mappings is\[{\left( n \right)^n}\]. Some important points to be keep in mind:

Two or more elements of A may have the same image in B.

f: x →y means that under the function of ‘f’ from A to B, an element x of A has image y in B.

It is necessary that every f image is in B but there may be some elements in B which are not f images of any element of A.

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