A mapping is selected at random from all the mappings defined on the set A consisting of three distinct elements. The probability that the mapping selected is one to one is

\frac{1}{9}

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\frac{1}{3}

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\frac{1}{4}

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\frac{2}{9}

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Solution

The correct option is B $\frac{2}{9}$
Let us consider a set A consisting of $3$ distinct elements.
$∴$ Total mapping $= 3 \times 3 = 9$
$\Rightarrow p$ (mapping selected is one to one ) $= \frac{2}{9} .$
Hence, the answer is $\frac{2}{9} .$

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A mapping is selected at random from all the mappings defined on the set A consisting of three distinct elements. The probability that the mapping selected is one to one is

The correct option is B 29Let us consider a set A consisting of 3 distinct elements.∴ Total mapping =3×3=9⇒p (mapping selected is one to one ) =29.Hence, the answer is 29.

The correct option is B $\frac{2}{9}$
Let us consider a set A consisting of $3$ distinct elements.
$∴$ Total mapping $= 3 \times 3 = 9$
$\Rightarrow p$ (mapping selected is one to one ) $= \frac{2}{9} .$
Hence, the answer is $\frac{2}{9} .$