Question

8. A plane is missing and is presumed to have probability of i/6 of being down in one of three regions, i=1,2,3. If a plane is actually down in region i, suppose there is a probability of 1- i/6 that the plane will be found upon a search of the i-th region. a) What is the probability the plane is in region 3, given a search of region 3 did not find the plane? b) What is the probability the plane is in region 1, given a search of region 2 did not find the plane?

Answer 1

Answers:-

(a)The probability that the plane is in region 3 and it is given that search of region 3 did not find the plane is P(R3/N3) is 9/11 or 0.81812.

(b)The probability that the plane is in region 1, given that the search of region 2 did not find the plane is P(R1/N2) is 1/8 or 0.125.

Solution-

According to question, the missing plane have probability of i/6 of being down in one of three regions, i=1,2,3.

Let it be denoted as P(Ri)

So, P(Ri)=i/6

Now,If a plane is actually down in region i, then the probability that the plane will be found upon a search of the i-th region is 1-i/6.

Let it be denoted as P(Fi/Ri)

So, P(Fi/Ri) =1- i/6

Now,If a plane is actually down in region i, then the probability that the plane will not be found upon a search of the i-th region is 1-(1-i/6).

Let it be denoted as P(Ni/Ri).

So, P(Ni/Ri) = 1- P(Fi/R) =1 -(1-i/6) = i/6

Or P(Ni/Ri)=i/6

Again, If a plane is actually down in one region, then the probability that the plane will not be found upon a search in another region is always 1.

Let it be denoted as P(Ni/Rj) i, j belongs to {1,2,3} and i≠j.

So, P(Ni/Rj) =1 for i≠j

Now,

(a) The probability that the plane is in region 3 and it is given that search of region 3 did not find the plane is P(R3/N3).

Now, by using Bayes' Theoram, we get

6 can be cancel out.So,

The probability that the plane is in region 3 and it is given that search of region 3 did not find the plane is P(R3/N3) is 9/11 or 0.81812

(b) The probability that the plane is in region 1, given that the search of region 2 did not find the plane is P(R1/N2)

Again, by using Bayes' Theoram, we get

6 can be cancel out.So,

Hence, the The probability that the plane is in region 1, given that the search of region 2 did not find the plane is P(R1/N2) is 1/8 or 0.125