I'm dealing with a sample problem where I want to work out the probability of a fair coin toss landing heads and a fair die roll landing 6. We are then told that at least one of those events has happened.
Why is the probability of this not as simple as P(C)P(D) = 0.083? How would one work out the correct probability using the Bayes theorem?
Thanks
On
It is clear that $P(C \cap D)= P(C)P(D)$. However in this case you are not asking that question , your question here is: $P(C \cap D \mid \text{ at least one of the events happened} )$ . In this new probability space the events are not independent. $$P(C \cap D \mid \text{ at least one of the events happened} )=P(C \cap D \cap \text{ at least one of the events happened} ) \frac{1 }{P(\text{ at least one of the events happened}) }$$ The first term is $0.083$, the second term is equal to $$ P(C \cup D)=P(C)+P(D)-P(C \cap D)$$ which is $\frac{1}{2} + \frac{1}{6}- 0.083$
Because now you have the additional information that at least one of those events has happened.
Hence the corresponding probability is $$\frac{P(CD)}{P(CD^c) + P(C^cD) + P(CD)}=\frac{P(CD)}{1-P(C^cD^c)}$$
That is I have excluded the possibility of $C^cD^c$ from happening.
Try to compute this quantity.