Question The air rescue service of the armed forces divided

Question: The air rescue service of the armed forces divided the path of a downed plane into search sectors. In one mission a search plane will overy every square kilometer of the designated sector. From past experience, one knows however that in one mission there is only a 50% chance of spotting a plane down in a sector of tundra. There is a 37.5% chance of spotting a plane down in a forested sector and a 12.5% chance of spotting a plane down in a lake sector.

(a) A small plane is down in an area containing two sectors of tundra, one lake sector and one forested sector and a priori the plane has an equal chance of being down in any one of the fours sectors. The mission director decides to use the rst mission to search the two sectors of tundra. What is the probability of spotting the downed plane?

(b) Given that we did not nd the plane in the tundra sections, what is the probability that the plane is (i) in the tundra sections? (ii) in the forested section? (iii) in the lake sector?

Solution

1. the plane is going over the two sectors of Tundra which has a probability of 50% of containing the downed plane. So P(T) . P(T) would be 0.5*0.5 = 0.25 is the probability of finding the plane in the two sectors of Tundra.

2. P(Finding Plane) would be the total probability of finding the plane in each sector so given that it has an equal chance of being in either sector is 2/4 * P(Tundra) + 1/4 * P(Forest) + 1/4*P(Lake)

(i)

(ii)

Pr(down in tundra |not found in tundra)=Pr(not found in tundra intersection down in tundra)/Pr(not found in tundra) =(Pr(down in tundra )Pr(not found in tundra|down in tundra))/Pr(not found in tundra))

=(((2/4)×(1/2)/(1-(1/4)))=1/3

because Pr(down in tundra )=2/4

  Pr(not found in tundra|down in tundra )=1/2

  Pr(not found in tundra)=1-Pr(found in tundra)=1-(1/4)

(ii)

Pr(down in Forest |not found in tundra)=Pr(not found in tundra intersection down in Forest )/Pr(not found in tundra) =(Pr(down in Forest )Pr(not found in tundra|down in Forest ))/Pr(not found in tundra))

=(((1/4)×(1)/(1-(1/4)))=1/3

because Pr(down in Forest )=1/4

  Pr(not found in tundra| down in Forest )=1

  Pr(not found in tundra)=1-Pr(found in tundra)=1-(1/4)

(iii)

Pr(down in lake|not found in tundra)=Pr(not found in tundra intersection down in lake)/Pr(not found in tundra) =(Pr(down in lake)Pr(not found in tundra|down in lake))/Pr(not found in tundra))

=(((1/4)×(1)/(1-(1/4)))=1/3

because Pr(down in lake)=1/4

  Pr(not found in tundra| down in lake)=1

  Pr(not found in tundra)=1-Pr(found in tundra)=1-(1/4)