Homework 3

Spatial Analysis, Spring 2001

Due Feb 6

1.

You are visiting a relative in Star Valley, who has lost eight of her prize hens (Gracie, Henrietta, Joline, Farah, Ming, Devi, Bobbi, and Ticabus) just as you came driving up. You help by catching three of the hens as they are trapped in an enclosure. How many possible combinations of three hens are there? What is the probability that your enclosure contains Henrietta, Ming, and Devi?

Answer: C₃⁸ = 56; probability is $\frac{1}{56}$

2.

It turns out that you actually did have those three hens in the enclosure. Now, you observe three more hens walking back into the coop. You make the arbitrary decision that they were first Jolene, then Ticabus, then Farah. What is the probability that you guessed correctly?

Answer: there are 5 chickens left. Order is specified, so the number of possible perumatations is P₃⁵ = 60; Probability is $\frac{1}{60}$

3.

Your relative wishes that you take a photograph of each of her eight hens posing next to each of three labrador retrievers that she owns (Larry, Curley, and Steve) and each of four antique clocks that she owns, so that each picture has one hen, one lab, and one clock. How many pictures is she asking that you take? How many of these pictures would have Henrietta in them?

Answer: $8 \times 4 \times 3 =$ 96 pictures. $4 \times 3 = 12$ pictures will have Henrietta.

4.

The chances of one of the hens bolting while you're taking these pictures is about $\frac{1}{5}$, and of the lab bolting about $\frac{1}{12}$. Assuming that these probabilities are statistically unrelated, what are the chances, while taking any one picture, of

(a)

both the chicken and the dog bolting?

Answer: $P(A \cap B) = \frac{1}{5} \times \frac{1}{12} = \frac{1}{60}$

(b)

either the chicken or the dog bolting?

Answer: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{5} + \frac{1}{12} - \frac{1}{60}= \frac{4}{15}$ or 0.2667

5.

If you're told that the dogs only bolt after a chicken has bolted, how would that change your answers to the previous question?

Answer A: $P(B \vert A) = \frac{1}{12}; \rightarrow P(A) \times P(B\vert A) = P(A \cap B) = \frac{1}{5} \times \frac{1}{12} = \frac{1}{60}$

Answer B: Since the dog only bolts after a chicken has bolted, then the probability of dog or chicken bolting is simply $\frac{1}{5}$