Laura’s diamond ring is missing and is presumed to have equal probability of bei

Question

Laura’s diamond ring is missing and is presumed to have equal probability of being accidently dropped in any of three shops she visited earlier in the day. If the ring is actually in shop 1 then the probability that the ring will be found upon a search of shop 1 is 0.23. If the ring is actually in shop 2 then the probability that the ring will be found upon a search of shop 2 is 0.51. If the ring is actually in shop 3 then the probability that the ring will be found upon a search of shop 3 is 0.26. What is the probability that the ring is in shop 1 given that the search of shop 1 was unsuccessful?

Explanation / Answer

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)

If A is made up of k mutually and collectively exhaustive sub-events, A1, A2,..Ak,

P(B) = sum over i = 1 to k of {P(B/Ai) x P(Ai)} ………………………………..(3)

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

Now to work out the solution,

Let events S1, S2 and S3 respectively represent the events that the ring had been accidently dropped in shop 1, 2 and 3.

Let F represent the event that the ring is found upon search.

With these representations, the given information translate into probability language as follows:

P(S1) = P(S2) = P(S3) = 1/3…………………………………………………………(1)

P(F/S1) = 0.23 …………………………………………………………………………(2)  

P(F/S2) = 0.51 …………………………………………………………………………(3)

P(F/S3) = 0.26 …………………………………………………………………………(4)

And, we want P(S1/FC)………………………………………………………………..(5)

Vide (4) of Back-up Theory, (5) = P(S1/FC) = P(FC/S1) x P(S1)/P(FC)…………..(6)

Again. P(FC) = 1 - P(F) and Vide (4) of Back-up Theory,

P(F) = {P(F/S1) x P(S1)} + {P(F/S2) x P(S2)} + {P(F/S3) x P(S3)}

= {(0.23)(1/3)} + {(0.51)(1/3)} +{(0.26)(1/3)}

= (1/3)(1)

= 1/3

Hence, P(FC) = 1 – (1/3) = 2/3.

Substituting in (6),

P(S1/FC) = P(FC/S1) x P(S1)/P(FC)

= [{1 - P(F/S1)} x (1/3)]/(2/3)

= (1 – 0.23)(1/2)

= (0.77)/2

= 0.385 ANSWER

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.