1) A hare and a coyote live in the same area. Within this area there is the wood

Question

1) A hare and a coyote live in the same area. Within this area there is the woods, the field and the lake. The coyote moves between the three areas each day with the following probabilities. The hare doesn't go down to the lake, he only moves between the field and the woods each day with the probabilities listed below. Coyote Hare W F L W F F L6 .4 L L0 .1 .9 If the hare and the coyote are in the same area at the same time the coyote eats the hare. Assume that the coyote starts at the lake and the hare starts in the woods What is the probability that the hare gets eaten in the woods? How many days is it expected to be before the hare gets eaten? a. b.

Explanation / Answer

a)

Coyote in woods probability given it starts at lake (Major portion)

= L-> F -> W , L-> L->F->W, L-> L-> L ->F->W and so on

= (0.1)(0.2) + (0.9)(0.1)(0.2) + (0.9)^2*(0.1)*(0.2)+..

{This is G.P. with r=0.9 and a= (0.1)*(0.2)}

= (0.1)*(0.2) / (1 -0.9) (S= a/(1-r) where r<1)

= 0.2

Hare in woods probability given it starts in woods = W + W->F->W

= 0.8 + 0.2*0.6 = 0.92

Thus, required probability roughly = 0.20/0.92 = 0.22

b)

For case a number of days required = 1/0.22 = 4.6 approx 5 days

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.