Question

WLD Incorporated, a national data-collection agency, estimates that 60% of all customers at home warehouse stores (in the United States) own their own home. WLD also estimates that 54% of all home warehouse customers have lived at their current address for less than five years, and that 71% of all home warehouse customers own their own home or have lived at their current address for less than five years (or both). Using these estimates, what is the probability that a randomly selected home warehouse customer both owns her own home and has lived at her current address for less than five years

Answer 1

Answer:

$P(W\ n\ F) = 43\%$

Step-by-step explanation:

Given

Represent customers that own their homes with W

Represent customers that have lived for less than 5 years with F

Such that:

$P(W) = 60\%$

$P(F) = 54\%$

$P(W\ or\ F) = 71\%$

Required

Determine $P(W\ n\ F)$

In Probability:

$P(W\ or\ F) = P(W) + P(F) - P(W\ n\ F)$

Substitute values for each

$71\% = 60\% + 54\% - P(W\ n\ F)$

$71\% = 114\% - P(W\ n\ F)$

Solve for P(W n F)

$P(W\ n\ F) = 114\% - 71\%$

$P(W\ n\ F) = 43\%$