Question

Based upon relative frequency, it has been determined that 15% of all cars tested emit excessive hydrocarbons, 12% emit excessive CO, and 8% emit excessive amounts of both. Let E and F denote the events that a randomly selected car emits excessive hydrocarbons and CO, respectively. Express the following events in terms of E and F and find the appropriate probabilities:
emissions of both hydrocarbons and CO are excessive

at least one emission is excessive

neither emission is excessive

hydrocarbon emission is not excessive

hydrocarbon emission is excessive, but CO emission is not

Answer 1

Answer:

P(F∩E)=8%

P(F∪E)=19%

P(F∪E)'=81%

P(E)'=85%

P(E)-P(E∩F)=7%

Step-by-step explanation:

If E denote the event that a randomly selected car emits excessive hydrocarbons and 15% of all cars tested emit excessive hydrocarbons, Then, the probability that a car emits excessive hydrocarbons is:

P(E) = 15%

At the same way, The probability that a car emits excessive CO is:

P(F) = 12%

Finally, if 8% emit excessive amounts of both, the probability that a car emits excessive hydrocarbons and CO is:

P(F∩E)=8%

it means that the probability that emissions of both hydrocarbons and CO are excessive is P(F∩E)=8%

On the other hand, the probability that at least one emission is excessive is the probability that the car emits excessive hydrocarbons or excessive CO and it is calculated as:

P(F∪E) = P(F) + P(E) - P(F∩E) = 15% + 12% - 8% = 19%

Additionally, The probability that neither emission is excessive is the complement of the last probability, so it is calculated as:

P(F∪E)' = 100% - P(F∪E) = 100% - 19% = 81 %

At the same way, the probability that hydrocarbon emission is not excessive is the complement of the probability of emit excessive hydrocarbon and it is calculated as:

P(E)' = 100% - P(E) = 100% - 15% = 85%

Finally, the probability that hydrocarbon emission is excessive, but CO emission is not is denoted and calculated as:

P(E) - P(E∩F) = 15% - 8% = 7%