Problem 2.2 Sherlock Holmes is trying to determine Professor Moriarty\'s recent

Question

Problem 2.2 Sherlock Holmes is trying to determine Professor Moriarty's recent whereabouts. His approach is to go to a random street and ask a random shopkeeper, "Has Professor Moriarty visited this street recently?" to which they will answer either "Yes" or "No." There is one problem: some of the streets are secretly controlled by Moriarty and the shopkeepers wil lie to protect him. The probability that a street is controlled by Moriarty is 1/2. The probability that Moriarty has visited a street recently is 1/4 (regardless of whether he controls it or not). If the street is controlled by Moriarty, the shopkeepers wil lie with probability (w.p.) 3/4. If the street is not controlled by Moriarty, a shopkeeper will lie w.p. 1/5. (a) Sherlock picks a random street and asks a shopkeeper if Moriarty has visited recently. Draw and label a tree diagram of all the probabilities and conditional probabilities leading up to

Explanation / Answer

Solution

As suggested in the question itself,

Let C represent the event that the street is controlled by Prof.Moriarty, V represent the event that Prof.Moriarty has visited, Y represent the event that the shop-keeper answers ‘yes’ and L represent the event that the shop-keeper tells lie.

Given P(C) = ½, P(V) = ¼, P(L/C) = ¾ and P(L/CC) = 1/5.

Back-up Theory

1. Multiplicative law: P(A B) = P(A) x P(B)

2. Conditional Probability: P(A/B) = P(A B)/P(B)

3. Bayes’ Theorem: P(A/B) = {P(B/A) x P(A)}/P(B)

4. P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}.    

Part (a)

Using the above theory and the given data, all possibilities along with the respective probability are presented in the table given below:

Shopkeeper’s response

C (½)

CC (½)

Total

V (¼)

VC (¾)

V (¼)

VC (¾)

(½ x ¼) = 1/8*1

(½ x ¾) = 3/8*1

(½ x ¼) = 1/8*1

(½ x ¾) = 3/8*1

Yes (Y)

(1/8)(¼) = 1/32

(3/8)(¾) = 9/32

(1/8)(4/5) = 1/10

(3/8)(1/5) = 3/40

39/80

No (YC)

(1/8)(¾) = 3/32

(3/8)(¼) = 3/32

(1/8)(1/5) = 1/40

(3/8)(4/5) = 3/10

72/160

Total

1/8

3/8

1/8

3/8

1

NOTE: *1- these are probabilities obtained by multiplying the corresponding event probabilities.

Probability of 1/32 under CVY is obtained as follows:

CVY represents the joint event that the street is controlled by Prof.Moriarty, that Prof.Moriarty has visited, and the shop-keeper answers ‘yes’.

This can happen only if the shop-keeper tells the truth for which the probability is ¼ since P(L/C) = ¾. Thus, the probability = ( ½ )( ¼ )( ¼ ) = 1/32.

All the other probabilities are obtained by following this very same concept. ANSWER

Part (b)

P[CC V Y] = 1/10 ANSWER [refer intersection Row4 and Column4]

Part (c)

P(Y/V) = (1/32) + (1/10) = 21/160 ANSWER [refer intersection Row4 and Column2 and intersection Row4 and Column4]

Part (d)

P(Y) = (1/32) + (9/32) + (1/10) + (3/40) = 38/80 ANSWER [sum of all probabilities in row 4

Shopkeeper’s response

C (½)

CC (½)

Total

V (¼)

VC (¾)

V (¼)

VC (¾)

(½ x ¼) = 1/8*1

(½ x ¾) = 3/8*1

(½ x ¼) = 1/8*1

(½ x ¾) = 3/8*1

Yes (Y)

(1/8)(¼) = 1/32

(3/8)(¾) = 9/32

(1/8)(4/5) = 1/10

(3/8)(1/5) = 3/40

39/80

No (YC)

(1/8)(¾) = 3/32

(3/8)(¼) = 3/32

(1/8)(1/5) = 1/40

(3/8)(4/5) = 3/10

72/160

Total

1/8

3/8

1/8

3/8

1

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