Answer:
(a) A y P(A) = 0.4 (b)  y
 y  =0.5 (c)
=0.5 (c)  ∪
∪ y P(
 y P( ∪
∪ ) = 0.9 (d)
) = 0.9 (d)  ∩
∩ y P(
 y P( ∩
∩ )=0.2
)=0.2
Step-by-step explanation:
A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So
P( ) = 0.3, P(
) = 0.3, P( ) = 0.4 and P(
) = 0.4 and P( ) = 0.1
) = 0.1
(a) P(A) = P( ) = P(
) = P( ) + P(
) + P( ) = 0.1 + 0.3 = 0.4
) = 0.1 + 0.3 = 0.4
(b)  P(B) = P( ) = P(
) = P( ) + P(
) + P( ) = 0.1 + 0.4 = 0.5
) = 0.1 + 0.4 = 0.5
P(  ) = 1-P(B) = 1-0.5 = 0.5
) = 1-P(B) = 1-0.5 = 0.5
(c) The customer does not buy from outlet 1 is the complement of A, i.e.,   , and the customer does not buy from outlet 2 is the complement of B, i.e.,
, and the customer does not buy from outlet 2 is the complement of B, i.e.,   , so, the customer does not buy from outlet 1 or does not buy from outlet 2 is
, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is   ∪
∪  and P(
 and P( ∪
∪  ) = P(
) = P( ) by De Morgan's laws
) by De Morgan's laws
P( )  = 1-P(A∩B)=1-0.1=0.9
)  = 1-P(A∩B)=1-0.1=0.9
(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to  ∩
∩ and P(
 and P(  ∩
∩ ) = P(
) = P( ) by De Morgan's laws, and
) by De Morgan's laws, and
P( ) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2
) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2