Answer:
The probability that Bill drove to work given that he reached home by 6:30 PM is 0.3324.
Step-by-step explanation:
Denote the events as follows:
T = Bill takes the train to work
D = Bill drives to work
X = Bill gets home from work by 6:30 PM.
The information provided is:
P (T) = 0.60
P (D) = 0.40
P (X|T) = 0.83
P (X|D) = 0.62
The Bayes's theorem states that the conditional probability of an event Ei given that another event has already occurred is:
[tex]P(E_{i}|X)=\frac{P(X|E_{i})P(E_{i})}{\sum\limits^{n}_{i=1}P(X|E_{i})P(E_{i})}[/tex]
Use the Bayes' theorem to determine the value of P (D|X) as follows:
[tex]P(D|X)=\frac{P(X|D)P(D)}{P(X|D)P(D)+P(X|T)P(T)}=\frac{(0.62\times 0.40)}{(0.62\times 0.40)+(0.83\times 0.60)}=0.3324[/tex]
Thus, the probability that Bill drove to work given that he reached home by 6:30 PM is 0.3324.
