Question

1. (17 points) In a pet friendly vicinity area 65.43% of households have at least one pet. In this vicinity, 51.51% of households are in the suburbs. 69.75% of these suburban households have at least one pet. a) (3 points) Let A denote the event that a household from this vicinity has at least one pet. Let B denote the event that a household from this vicinity is in the suburbs. Collect and organize into proper probability notation all of the above information about a randomly selected household from this vicinity. b) (4 points) What is the probability that a randomly selected household from this vicinity is in the suburbs and has at least one pet? c) (2 points) If a randomly selected household from this vicinity is in the suburbs, what is the probability that it does not have any pet? d) (4 points) Among the households in this vicinity, are the events being in the suburbs and having at least one pet, independent? Justify your answer with probabilities. e) (4 points) If a randomly selected household in this vicinity has at least one pet, what is the probability that the household is in the suburbs?

Answer 1

solution :- Da) PCA) = 0.6543 P (B) = 0.5151 PLA) : 0.6975 b P(ANB) = P1A). PIB) = 0.6975*0.515) 10.3593 9P1031-P%) = 1-0-6975 10.3025 -0.6975 .302

0.769754B | not independent d) P(A) = 0.6975 P (A) = 0.6543 so, ADB d) P (A) = P(ANB) - P(A) = 0.6975% O. SISI 0.65 43 = 10.54917 0.5491