Q1a If something goes wrong, your friend will not make it on time, so PD|-X)=1 a

Question

Q1a If something goes wrong, your friend will not make it on time, so PD|-X)=1 and P(D|X)=0.5

10 mark(s)] You invite your friend to come to your house for dinner. They have to drive from a fair distance away, so the probability distribution of their arrival time t is uniform between 7pm and 8pm, assuming everything is going to plan. At 7:30pm, you start to worry because your friend has not arrived. Let D1 be the statement that your friend did not arrive before 7:30pm. Let X be the statement that everything is fine, so -X is the statement that something is wrong. (a) [2 mark(s)) Find the probabilities P(Di | X) and P(Di |-X). As diseussed in lectures, these relative frequencies or proportions are not the same concept as the Bayesian probabilities needed for the question, but in some applications it is sensible to set them to the same values.

Explanation / Answer

(a)

If something goes wrong, your friend will not make it on time, so PD1|-X)=1 and P(D1|X)=0.5

(b)

Given P(X) = 0.99, P(-X) = 0.01

By law of total probability,

P(D1) = P(D1 | -X) P(-X) + P(D1 | X) P(X)

= 1 * 0.01 + 0.5 * 0.99 = 0.505

By Bayes rule,

P(X | D1) = P(D1 | X) P(X) / P(D1) = 0.5 * 0.99 / 0.505 = 0.98

P(-X | D1) = P(D1 | -X) P(-X) / P(D1) = 1 * 0.01 / 0.505 = 0.02

(c)

If something goes wrong, your friend will not make it on time, so PD2|-X)=1

and P(D2|X)=1 - 54/60 = 0.1 (arrival time is a uniform ditribution variable)

By law of total probability,

P(D2) = P(D2 | -X) P(-X) + P(D2 | X) P(X)

= 1 * 0.01 + 0.1 * 0.99 = 0.109

We know that if D2 happens, the probability of D1 happens will be 1, So P(X | D1, D2) = P(X | D2)

By Bayes rule,

P(X | D1, D2) = P(X | D2) = P(D2 | X) P(X) / P(D2) = 0.1 * 0.99 / 0.109 = 0.9083

P(-X | D1, D2) = P(-X | D2) = P(D2 | -X) P(-X) / P(D2) = 1 * 0.01 / 0.109 = 0.0917

d)

Now the prior probabilities be,

P(X | D1) = 0.98

P(-X | D1) = 0.02

Let A be X | D1 and then -A is -X | D1

amd P(A) = 0.98 and P(-A) = 0.02

If something goes wrong, your friend will not make it on time, so PD2|-A)= 1

and P(D2 | A)=1 - 24/30 = 0.2 (arrival time is a uniform ditribution variable)

P(D2) = P(D2 | -A) P(-A) + P(D2 | A) P(A)

= 1 * 0.02 + 0.2 * 0.98 = 0.216

By Bayes rule,

P(A | D2) = P(D2 | A) P(A) / P(D2) = 0.2 * 0.98 / 0.216 = 0.9074

P(-A | D2) = P(D2 | -A) P(-A) / P(D2) = 1 * 0.02 / 0.216 = 0.0926

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