The probability that Chloe is happy given that she has had dinner by 6pm is 0.9. The probability that Chloe is not happy given that she has not had dinner by 6pm is 0.8. Assume that there is a 60% chance that Chloe eats dinner by 6pm on a given day. a) If Chloe is not happy at 6pm, find the probability that she has not had dinner yet. b) If Chloe is happy at 6pm, find the probability that she has already had dinner.

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

Step-by-step 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.

150.6

all i did was i multiplied 5.02 by 30

no

step-by-step explanation:

answer