A student went to the Olympic peninsula to study crow nesting in Douglas fir trees. In one particular plot she sampled 100 trees. She recorded the number of trees above and below 30 meters in height, as well as those with and without nests. In her sample she counted 22 trees below 30 m in height that had no nests and 35 trees greater than 30 m in height that had one or more nests. In total there were 38 trees less then 30 m tall and 62 trees greater than 30 m tall.

HINT: To answer these questions try organizing the data in a contingency table.

a) If I am standing under a tree less than 30 m tall, what is the probability I find a crow’s nest? What about greater than 30 m tall? Based on these probabilities, if you want to find a crow’s nest, are you better off looking at tall or short trees?

P(nest|short) = 42.1%

P(nest|tall) = 56.5%

Given these probabilities, you’re better off looking at taller trees

b) What is the probability a tree is greater than 30 m tall or it has a crow’s nest?

P(nest or tall) = 0.62+0.51-0.35 = 0.78 = 78%

c) Is the tree in which a crow nests independent of tree height (i.e. tall (>30m) vs. short (<30 m))? In general you can answer this by showing whether being greater than 30 m tall and having a nest are independent.

P(tall) = 0.62 = 62%

P(tall|nest) = 0.686 = 68.6%, not equal so not independent

d) What is the probability that the first two trees I look at have one or more crow’s nests?

P(nest) = 0.51, P(nest.nest) = 0.51² = 0.26

(this assumes that the two findings of nests are independent, i.e.

that finding the first nest doesn’t change the chances of finding the

second.

At a particular research site involved in the study of brown bears and grizzly bears, there were 24 local bears that the researchers observed and trapped on a regular basis. There were 16 grizzly bears and 8 brown bears. Of the grizzly bears, 12 were male and 4 were female. Of the brown bears, 4 were male and 4 were female.

Since bears are difficult and dangerous to work with, the researchers would set one trap in a different location every afternoon and find a single bear captured every morning. After a brief examination and weighing, the bear was released. Assume that every bear is equally likely to be caught on a given day.

HINT: To answer these questions, try organizing the data in a contingency table

a) While walking out to inspect the trap, one of the researchers can tell from a distance that today he has captured a grizzly bear. What is the probability that the bear is male?

P(male|grizzly) = 12/16 = 0.75

b) What is the probability of going three days without capturing a female bear?

P(no female in 3 days) = P(male,male,male) = (16/24)³ = 0.30

c) What is the probability of capturing the same individual bear (they all have ear tags, so can be uniquely identified) two days in a row?

P(same bear two days) = (1/24)² = 0.0017, assuming each day is independent

d) A relocation program is started, so now every bear captured is sedated and flown far away (removing them from the local population). What is the probability that the first two bears to be evacuated are both grizzly bear males?

(12/24) * (11/23) = 132/552 = 0.239

In an experiment to test the susceptibility of an endangered plant species to fungus, some specimens of the plant were raised, along with a related exotic species and marigolds (as a control) in a greenhouse. At two months of age they were inoculated with the fungus, and one month later all of the plants were classified as healthy, diseased, or dead. The results are reported in the contingency table below:

	Control	Endangered	Exotic	Totals
Dead	15	16	6
Diseased	21	3	9
Healthy	24	0	42
Totals				136

a) List three variable pairs that are mutually exclusive.

{contr,end}, {contr,exot}, {exot,end}, {dead,dis}, {dis,heal}, {heal,dead}

a) If a plant is selected from the greenhouse at random, what is the probability that it is either dead or diseased?

37/136 + 33/136 + 0/136 = 0.515

b) What is the probability it is exotic and dead?

57/136 * 6/57 = 0.0441

c) Are endangered plants independent of healthy plants?

P(end) = 19/136

P(end|heal) = 0, therefore NOT independent