Think I've managed to get it but not 100% sure
The probability that Ted is late for work on a rainy day is 0.4 and the probability that he is late for work on a non-rainy day is 0.1. The probability that it will rain tomorrow is 0.6. What is the probability that it did not rain if Eddy was late at work?
I've said $P($Late given it rains)$=0.4$ $P($Late given it doesn't rains)$=0.1$ $P$(Rains tommorow)$= 0.6$
I've then worked out $P$(Late on any day)$= 0.28$ $P$(no rain given late)$= ($P(Late given no rain)*P(no rain))/P(late) = $(0.4*0.1)/(0.28)$ = 1/7
Correct?
To give an answer to question I will outline the method
We want to compute the following, $$ P(\text{No Rain}|\text{Late}) $$ What information do we have $$ P(\text{Late}|\text{Rain}) = 0.4,\\ P(\text{Late}|\text{No Rain}) = 0.1,\\ P(\text{Rain}) = 0.6 \implies P(\text{No Rain}) = 1 - P(\text{Rain}) = 0.4 $$ We can use Bayes to define the following $$ P(\text{No Rain}|\text{Late})P(\text{Late}) = P(\text{Late}|\text{No Rain})P(\text{No Rain}) $$ or $$ P(\text{No Rain}|\text{Late})= \frac{P(\text{Late}|\text{No Rain})P(\text{No Rain})}{P(\text{Late}) } $$ The only information we are missing is $P(\text{Late})$ This is computed by $$ P(\text{Late}) = P(\text{Late}|\text{No Rain})P(\text{No Rain}) + P(\text{Late}|\text{Rain})P(\text{Rain}) $$ we are no ready $$ P(\text{Late}) = 0.1 \times 0.4 + 0.4\times0.6 = \frac{28}{100} $$ Now putting it int o the equation we want $$ P(\text{No Rain}|\text{Late})= \frac{\frac{1}{10}\times\frac{4}{10}}{\frac{28}{100}} = \frac{4}{28} = \frac{1}{7} $$ So the answer is correct.