Answer to Question #202347 in Statistics and Probability for Aasia Tariq

Question

In a small town, two lawn companies fertilize lawns during the summer. Tri-State Lawn Service has 72% of the market. Thirty percent of the lawns fertilized by Tri-State could be rated as very healthy one month after service. Greenchem has the other 28% of the market. Twenty percent of the lawns fertilized by Greenchem could be rated as very healthy one month after service. A lawn that has been treated with fertilizer by one of these companies within the last month is selected randomly. If the lawn is rated as very healthy, what are the revised probabilities that Tri-State or Greenchem treated the lawn?

Answer 1

Let T represent Tri-State Lawn, G represent Greenchem, and H represent Healthy after one month.

$P(T)=0.72$

$P(G)=0.28$

$P(H| T)=0.3$

$P(H| G)=0.2$

$P(T|H)$

$P(G|H)$

Since T and G are mutually exclusive ( both firms cannot fertilize the same lawn at the same time), $P(T\cup G)=P(T)+P(G)$ which is equal to 1 in this case (0.72+0.28) since the events are exhaustive.

Thus, $P((T\cup G)|H)=P(T|H)+P(G|H)=1$

But,

$P(T|H)=\frac{P(H|T)P(T)}{P(H|T)P(T)+P(H|G)P(G)}$

$=\frac{0.3×0.72}{(0.3×0.72)×(0.2×0.28)}=\frac{0.216}{0.272}$

$=0.794$

Similarly,

$P(G|H)=\frac{P(H|G)P(G)}{P(H|G)P(G)+P(H|T)P(T)}$

$=\frac{0.2×0.28}{(0.2×0.28)×(0.3×0.72)}=\frac{0.056}{0.272}$

$=0.206$