12. Uncle Ra and Aunt Heather took nephew Jonathan and niece Bridgit to Spain. W

Question

12. Uncle Ra and Aunt Heather took nephew Jonathan and niece Bridgit to Spain. While riding the bullet train from Madrid to Cordoba it seemed to them that castles appeared on the horizon in a rather random fashion, but with an intensity of about four per hour. (Break the total time period into disjoint one hour intervals.) (a) Build a mathematical model for this situation, and then answer the following questions in the context of your model. (b) What is the probability that they will see at most two castles in the first hour? (*) Additionally, suppose that the two sibs are equally good at being the first to sight the castles that do appear. (c) What is the probability that Jonathan will sight at least 4 of the first 8 castles? d) What is the probability that Jonathan will be the first to site at least 3 castles uring one of the hour long intervalsjgWe^eteneef, then does. (e) What is the probability that Jonathan will sight 4 before Bridget sights 3? (f) Reevaluate this probability assuming that Jonathan is twice as good as Bridget in the matter of sighting castles. (8) Criticize the model you developed in part (a).

Explanation / Answer

The number of castles that will arrive in time t ~ P(4 t)

The time between arrivals is distributed E(4), exponential with parameter 4.

b) In one hour, the castles appearing will be distributed P(4*1) = P(4)

P(X <= 2) = P(0) + P(1) + P(2) = e^-4 + e^-4*4^1/1! + e^-4*4^2/2! = e^-4(1 + 4 + 8) = 13e^-4 = 0.238103305553544

c) Since they are equally good, p = 1/2

d) P(Jonathan will sight at least 4 castles)   We can either add P(4)+P(5)+P(6)+P(7)+P(8) or 1 - P(0-3) = 1 - P(X <= 3) =

1 - binomdist(3, 8, 1/2, TRUE)

Another way to solve this:

Note that each is equally likely to see 0-3 and 5-8 castles. Thus, 1/2 = P(5-8) + 1/2P(4)

Thus, P(Jonathan will see 4 or more) = 1/2 + 1/2 P(4) = 1/2 + 1/2 C(8,4)/2^8 = 1/2+1/2*70/256 =

1/2+35/256 = 163/256 = 0.63671875

e) The distribution for Jonathan sighting castles first in one hour ~ P(4 * 1 * 1/2) ~ P(2)

Then, the probability that he is the first to see at least 3 is 1 - P(0) - P(1) - P(2) = 1 - ( e^-2 + e^-2*2^1/1! + e^-2*2^2/2!) =

1 - 5 e^-2 = 0.3233

f) P(Jonathan sights 4 before Bridget sights 3 =

P(4 or more of the first six selected are seen by Jonathan (viewing it this way makes the analysis easier) = 1 - B(3; 6, 1/2)

We can solve this very similarly to how we solved d)

1 - (1/2 + P(3)) = 1/2 - 1/2 P(3) = 1/2 - 1/2 C(6,3)/2^6 = 1/2 - 1/2*20/2^6 = 1/2 - 10/2^6 = 11/32 = 0.3438

Of course, we could have simply added P(4) + P(5) + P(6), calculated 1 - binomdist(3,6,1/2,TRUE) or, noting that

P(4) + P(5) + P(6) = P(0) + P(1) + P(2), simply calculated binomdist(2,6,1/2,TRUE)

g) As castles were designed to protect probably a certain area, you would expect them to tend not to be as close together as you would expect at random.

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