Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is :
Solution
g(3) = 2g(1) can be defined in 3 ways<br><br>number of onto functions in this condition = 3 $\times$ 4!<br><br>Total number of onto functions = 6!<br><br>Required probability = ${{3 \times 4!} \over {6!}} = {1 \over {10}}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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