Events:
A=from store A
B=from store B
D=defective
then probability of defective player given it is from store A:
P(D|A)=P(D∩A)/P(A)=P(D∩A)/0.3=0.027
then probability of defective player given it is from store B:
Solve for P(D∩A)
P(D|B)=P(D∩B)/P(B)=P(D∩B)/0.7=0.037
Solve for P(D∩B)
Probability that a player is defective from any store
=P(D)
=P(D∩(A∪B))
=P(D∩A)+P(D∩B)
=sum of two values calculated above
Probability that a player is from store A, given that it is defective
P(A|D)
=P(A∩D)/P(D)
=P(D∩A)/P(D)
= evaluate using known values calculated above.
(hint: the answer is between 0.2 and 0.25)
A certain type of dvd player is sold at just two stores. 30% of sales are from store A and 70% of sales are from store B. 2.7% of the dvd players sold at store A are defective while 3.7% of the dvd players sold at store B are defective. If Kate receives one of these dvd players as a gift and finds that it is defective, what is the probability that is came from store A?
1 answer