Question

If a child is given cards with A, C, D, G, O and T on them, what is the probability that he/she could spell DOG by the correct arrangement of 3 cards from 6?

Answer 1

There are $A_6^3=\frac{6!}{3!}=4\cdot5\cdot6=120$ different ways of choosing the word that has 3 letters from the set of 6 letters. The probability to receive the word DOG is $\frac{1}{120}=0.0083$ (it is rounded to 4 decimal places)

Answer: 0.0083 (the answer is rounded to 4 decimal places)