A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Question

I have this question as an example in my maths school book. The solution given there is:-

E = the man reports six

P(S1)= Probability that six actually occurs = $\frac{1}{6}$

P(S2)= Probability that six doesn't occur= $\frac{5}{6}$

P(E|S1)= Probability that the man reports six when six has actually occurred = $\frac{3}{4}$

P(E|S2)= Probability that the man reports six when six has not occurred = $1-\frac 3 4=\frac 1 4$

Therefore, by Bayes' Theorem,
$P(S1|E)=\frac{(\frac{1}{6}\cdot\frac{3}{4})}{(\frac{1}{6}\cdot \frac{3}{4})+(\frac{5}{6}\cdot \frac{1}{4})} =\frac{3}{8} $

I have its solution but my teacher said that the solution given is incorrect and told that the actual solution would be something else:-

$P(S1|E)=\frac{\frac 1 6\cdot\frac3 4}{(\frac 1 6\cdot \frac 3 4)+(\frac 5 6\cdot \frac1 4\cdot\frac1 5)} = \frac 3 4$

So, I want to ask which one is correct. Thank you.

Answer 1

The difference in solutions comes in the estimation of the probability that the man reports six when six has not occurred.

If the man randomly chooses a number to report when he lies (which seems like a reasonable statement), then the probability he chooses 6 is 1/5. If you multiply your calculation of P(E|S2) by this, you get your teacher's solution.