$F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}$
$F = \frac{30}{36} = \frac{5}{6}$
$E =$ The event that at least one lands on 6
$E = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}$
$E = \frac{11}{36}$
What is the probability of E, when given that F has occurred?
$P = \frac{P(EF)}{P(F)}$
$P = \frac{1-P(EF^c)}{P(F)}$
$P = \frac{\frac{11}{36}}{\frac{5}{6}}$
$P = \frac{11}{30}$
This answer was incorrect it is $\frac{1}{3}$ I think there is an issue in identifying $P(EF)$, what should that be and why? Identifying F is easy because that is what is given to have occurred.
Added: Is $EF$ equivalent to $E \cap F$?
Best Answer
The mistake in your work is that while the question asks us to find that the numbers on the dice are different, in writing the sample space for event $E $, you have also included $(6,6) $ where the numbers are not different. Correct that and we will get, $$P=\frac{\frac{10}{36}}{\frac {5}{6}} =\frac {1}{3} $$ Hope it helps.