Question

A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy

Answer 1

Answer:

(a) A y P(A) = 0.4 (b) $\bar{B}$ y $P(\bar{B})$=0.5 (c) $\bar{A}$∪$\bar{B}$ y P($\bar{A}$∪$\bar{B}$) = 0.9 (d) $\bar{A}$∩$\bar{B}$ y P($\bar{A}$∩$\bar{B}$)=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($A\cap \bar{B}$) = 0.3, P($B\cap \bar{A}$) = 0.4 and P($A\cap B$) = 0.1

(a) P(A) = P($A\cap (B\cup\bar{B})$) = P($A\cap B$) + P($A\cap \bar{B}$) = 0.1 + 0.3 = 0.4

(b)  P(B) = P($B\cap (A\cup\bar{A})$) = P($B\cap A$) + P($B\cap \bar{A}$) = 0.1 + 0.4 = 0.5

P( $\bar{B}$) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e.,  $\bar{A}$, and the customer does not buy from outlet 2 is the complement of B, i.e.,  $\bar{B}$, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is  $\bar{A}$∪ $\bar{B}$ and P($\bar{A}$∪ $\bar{B}$) = P($(A\cap B)^{c}$) by De Morgan's laws

P($(A\cap B)^{c}$)  = 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 $\bar{A}$∩$\bar{B}$ and P( $\bar{A}$∩$\bar{B}$) = P($(AUB)^{c}$) by De Morgan's laws, and

P($(AUB)^{c}$) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2