Artificial Intelligence

I need answer to question 4

The Pirate Treasure You are a seasoned tomb raider and have spent the last week rummaging through an old pirate cove full of treasure. So far you have opened 100 chests and of those, 50 have in fact contained treasure! Out of these 50, 40 were trapped and you sustained some painful damage from opening them. Out of these 40 trapped chests with treasure, 28 were also locked. Now, of the 10 untrapped chests with treasure, three were locked. One would think that only chests with treasure would be trapped, but these pirates were truly nasty, they also put traps on chests with no treasure. Of the 50 chests containing no treasure, 20 were trapped! You forgot how many of the chests without treasure were locked, but you believe that the ratios were similar to the ones with treasure. You have now discovered a new chest that you havent seen before. When you take a careful look, you notice that it is locked. What is the chance that this chest will contain treasure? What is the chance that it will be trapped? You are not feeling so good after all the previous traps, so will it be worth opening this chest if your life is on the line?

Question 1 (36 points) Construct a Bayesian Network that helps you answer these questions. Write down the propositions (nodes), the connections between them and the probability table for each node. You should be able to do this by essentially just reading the required information from the hints in the text (i.e., without any complicated computations).

Answer 1:

Probability tables:

Treasure	0.5
Treasure	0.5

	Trapped	Trapped
Treasure	0.8	0.2
Treasure	0.4	0.6

	Locked	Locked
Trapped	0.7	0.3
Trapped	0.3	0.7

P(Treasure)=50100 = 0.5

P(Trapped) =(40+20)100 = 0.6

P(Locked) = (28+3+14+9)100= 0.54

P(Trapped |Treasure) = 4050= 0.8

P(Trapped |Treasure) =2050 = 0.4

P(Locked |Trapped) = 28+1460=0.7

P(Locked|Trapped) =3+940= 0.3

Question 2 What is the probability that a locked chest contains treasure?

Answer 2:

To know the probability of a locked chest that contains treasure, its first good to know the probabilities of P(Locked | Treasure) and P(Locked | Treasure).

P(Locked | Treasure) = 28+3 /50 = 0.62

P(Locked | Treasure)= 14+9 /50 = 0.46

Formula for Bayes Theorem: P(A | B) = P(B | A) P(A) / P(B)

So, to calculate how many locked chests contains treasure, we can use the following formula:

P(Treasure | Locked) = P(Locked | Treasure) P(Treasure) / P(Locked)

P(Treasure | Locked) = 0.62 0.5 / 0.54 = 0.57 Hence, the probability that a locked chest contains treasure is 57%.

Question 3 What is the probability that a locked chest is trapped?

Answer 3:

To calculate the probability that a locked chest is trapped, we can use the following formula:

P(Trapped | Locked) = P(Locked | Trapped) P(Trapped) / P(Locked)

P(Trapped | Locked) = 0.7 0.60.54= 0.78 Hence, the probability that a locked chest is trapped is 78%.

Question 4 Develop a formula for computing the probability of a chest containing treasure, given that you observe whether it is locked or not. Write down the steps you took to get to the formula and the final formula. The formula must only use probabilities that are given in the Bayesian Network. Check whether your formula (and/or the Bayesian Network) is correct by comparing the values you compute with the formula with those that you get from the simulator. Note: The goal for this question is to look at how the computations in the previous questions can be done using the Bayesian Network. To solve this, first write down what you want to compute, then use use Bayes Law, Marginalization and the conditional independence rules to relate this to the given probabilities.

This could be helpful:

Laws of Probabilities The following laws of probabilities might be helpful for developing formulas: P (A) = 1 P (A) Bayes Law: P (A B) = P (A|B) P (B) Marginalization/Summing out: P (A) = P bP (A b)(= P (A B) + P (A B)) Independence 1: P (A|B, C) = P (A|C), if A and B are independent given C. Independence 2: P (A B|C) = P (A|C) P (B|C), if A and B are independent given C. (This follows from Independence 1 and Bayes Law.) All those laws allow arbitrary additional conditions as long as all probabilities have the condi- tion, for example: Bayes Law: P (A B|C) = P (A|B, C) P (B|C) Marginalization/Summing out: P (A|C) = P bP (A b|C)