A man is known to speak the truth 3 out of 5 times. He throws a die, and reports that it is a 1. Find the probability that it is actually 1. (CBSE 2014)
Solution 1
Assume the die is fair. If the man throws the die 30 times, the expected outcome is each of {1,2,3,4,5,6} occurs 5 times. He $$\text{tells the truth } \frac{3}{5} \times30=18 \text{ times}, \text{lies} \frac{2}{5}\times 30=12 \text{times}$$
Tabulating, he gets a 1 five times. Out of those, he tells the truth 3 times. Out of the 25 times he does not get a 1, he lies 10 times. Those lies are equally distributed among the possible lies, and hence he lies that it is a "1" two times.
Apply the logic of Bayes Theorem. There are five outcomes with a 1, out of which 3 are represent what we are looking for. Hence, the final probability is 3/5.
Says it is $\frac{3}{13}$
The solutions are different, and the difference of course comes from the different assumptions being made here. Which assumption/solution is correct, and why? Or is it that the question needs to be made more precise?