Question

A town has two malls. 62% of the town’s residents shop at Willow Creek Mall. 73% of the residents shop at Two Harbors Mall. 48% of the residents visit both malls. On any given day, what is the probability that a person chosen at random will visit either Willow Creek or Two Harbors?
A. 0.87
B. 0.64
C. 0.62
D. 0.75
E. 0.48

Answer 1

Answer: A. 0.87

Step-by-step explanation:

Let A be the probability that the town’s residents shop at Willow Creek Mall and B be the probability that the residents shop at Two Harbors Mall.

Then , we have

$P(A)=62\%=0.62\\\\\ P(B)=73\%=0.73\\\\\ \text{P(A and B)}=48\%=0.48$

Now,

$\text{P(A or B)=P(A+P(B)-P(A and B)}\\\\\Rightarrow\ \text{P(A or B)}=0.62+0.73-0.48=0.87$

Hence, the probability that a person chosen at random will visit either Willow Creek or Two Harbors = 0.87

Answer 2

We are given with the probabilities of visiting separately two malls and thr probability of visiting both malls. The probability of visiting Willow Creek or Two Harbors is equal to the sum of the individual probability of two malls minus the probability of visiting both malls. This is equal to 0.62 + 0.73 - 0.48. The answer here is A. 0.87.