I am pretty sure this is a simple question regarding Bayes' Rule, but just wanted to double check as the hint given is confusing me.
In a certain group, 40% of people read the New York Times. Of the NYT readers, 55% have an online subscription. Furthermore, 65% of the group closely follow NFL football, but only 12% of these football fans have an online subscription to the NYT.
I must answer the following questions:
(a) Given that a person does not have an online subscription, what is the probability they closely follow NFL football?
I have answered this is the following way, which I think is correct:
P(NFL | no subscription) = P(NFL & no subscription)/P(no subscription)
- P(no subscription) = People who read NYT who don't have subscription + people who don't read NYT (since we assume that if they dont read the NYT they will not have an online subscription) = (0.40 * 0.45) + 0.6 = 0.78
- I'm just unsure on how to find the numerator of the conditional probability equation..
(b) Given that an individual does not follow the NFL, what is the probability that they read the NYT? It will be helpful to assume that reading the NYT and watching NFL are conditionally independent.
- Any help on this part would be much appreciated.
Just apply the definion of conditional probability again.
$\Bbb P(\text{NFL } \& \text{ no subscription}) ~{= \Bbb P(\text{NFL})\, \Bbb P(\text{no subscription}\mid \text{NFL}) }$
You have these probabilities.
If you assume (conditional given the group) independence between newspaper readership and football followship, then by definition of independence:
$\begin{align}\Bbb P(\text{NFL & NYT}) ~&=~ \Bbb P(\text{NFL})\, P(\text{NYT})\\ \Bbb P(\text{NYT}\mid\text{NFL}) ~&=~ \Bbb P(\text{NYT})\end{align}$