Answer:
a) If Chloe is not happy at 6pm, the probability that she has not had dinner yet =P(B/D)= 0.8
b) If Chloe is happy at 6pm,the probability that she has already had dinner= P(A/C) = 0.9
Explanation:
Let A be an event that Chloe is happy given that she has had dinner by 6pm
P(A) = 0.9
Let B be an event that Chloe is not happy given that she has not had dinner by 6pm
P(B) = 0.8.
Let C be an event that Chloe eats dinner by 6pm on a given day
P(C) = 60%=0.6
Let D be an event that Chloe does not eat dinner by 6pm on a given day
P(D) = 1- 60%=1- 0.6= 0.4
This requires conditional probability .
Part a
P(B/D)=P(B∩D)/P(D)
P(B∩D) = 0.8 *0.4=0.32
P(D) = 0.4
P(B/D)=P(B∩D)/P(D)
= 0.32/0.4= 0.8
P(B)= P(B/D)
0.8 is equal to 0.8
This shows that the events are independent.
b)
P(A/C)= P(A∩C)/ P (C) = 0.9*0.6/0.6= 0.9
Which is again equal to probability of event A showing that the events are independent.