1.If P(A) = .46 and P(B) = .17 and P(A U B) = .63, then A and B are:
Select one:
a. mutually exclusive
b. collectively exhaustive
c. statistically independent
d. mutually exclusive and collectively exhaustive
e. none of the above/can’t be determined with info given
- Assume that the events C and D are mutually exclusive and the P(C U D)=.70. If the probability that D occurs is .45, what is the probability of C occurring?
2. Assume that events E and F are statistically independent and P(E)=.29 and P(F)=.37. What is P(E U F)?
3. Assume that events G and H are collectively exhaustive and P(G)=.7 and P(H)=.6. What is P(G ∩ H)?
1.A particular Land Trust Organization is collecting information about the use of land in a particular county and whether it is conserved for environmental purposes and the degree of that conservation. GAP status codes classify areas into highly protected, moderately protected (conserved but mainly used for other purposes), and unprotected. From the last survey, we know that 67% of the county is unprotected and 18% is highly protected. From the highly protected areas, 88% is forest and the rest is equally divided between urban use and agriculture use. Most of the unprotected land is used for urban purposes–85%. Twelve percent of what is moderately protected is forest. The rest of the moderately protected land is equally used for agriculture and urban purposes. We also know that the total amount of land used for agriculture in the county is 14%. Use four decimal places in the chart for this problem.
a. What is the probability that a parcel is for agriculture and is moderately protected?
b.If we are interested in forest conservation, what is the probability that a forest parcel is highly protected?
c.What is the probability of having a highly protected area or forest?
d.Given that the parcel is somehow protected (either highly protected or moderately), what is the probability that this parcel is used for agriculture or is an urban area?
1. Multiple choice quiz.:Select one:a. mutually exclusive probability of C1. P(C)=P(C U D)-P(D)=0.70-0.45=0.25 2.=1-[(1-P(E)*(1-P(F)]=1-(1-0.29)*(1-0.37)=1-(0.71*0.63)=1-0.4473=0.55273….
