Question

A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy
products from outlet 1 alone, 40% buy from outlet 2 alone, and 10% buy from both 1 and 2. Let A
denote the event that a potential customer, randomly chosen, buys from outlet 1, and let B denote
the event that the customer buys from outlet 2. Suppose a potential customer is chosen at random.
For each of the following events, represent the event symbolically and then find its probability.

a) The customer buys from outlet 1.

b) The customer does not buy from outlet 2.

c) The customer does not buy from outlet 1 or does not buy from outlet 2.

d) The customer does not buy from outlet 1 and does not buy from outlet 2.

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