Question

Let \[ S=\{2,3,5,7,11,13\} \] Consider all onto functions from \(S\) to \(S\). If function \(f\) is chosen randomly, probability that \[ f(3)>3f(2) \] is

• A. \(\frac23\)
• B. \(\frac2{15}\)
• C. \(\frac16\)
• D. \(\frac1{10}\)

Hint:

If domain and codomain have same number of elements, onto automatically means bijection.

Answer 1

The Correct Option is D

Concept: Since domain and codomain have equal number of elements, onto function means bijection. Total functions: \[ 6! \]

Step 1: Count favorable assignments.
Condition \[ f(3)>3f(2) \] Possible values set: \[ 2,3,5,7,11,13 \] Choose ordered pairs satisfying condition. Possible pairs: \[ (2,7),(2,11),(2,13) \] \[ (3,11),(3,13) \] \[ (5, ) \] Total valid ordered pairs = 6.

Step 2: Arrange remaining values.
Remaining elements can permute in \[ 4! \] Thus favorable functions \[ 6\times4! \]

Step 3: Probability.
\[ P= \frac{6\times4!}{6!} \] \[ = \frac6{30} \] \[ = \frac1{10} \] Hence \[ \boxed{\frac1{10}} \]