Question

Rob (who is a robot roaming in the basement of the new CSE building) has lost his graphics card. He is now unable to amuse himself playing video games at the times when he is not picking up empty coke cans or escorting lost students out of the basement. He has lost the card in either room A (with probability 1/3) or room B (with probability 2/3). These rooms are humid and rat infested and the probability that the card (which was good and working just before it was lost) will survive a night in room A is 4/5. Room B is even worse: the probability that the graphics card is still good after a night in room B given that it was good the day before is 3/5. Call the event of the card being in good working condition G If the card is in room A and Rob spends a day searching for it in A then the probability of finding it (call this event F) that day is 1/2. The similar probability for finding the card on a day of search in room B is 2/5. Assume that the card surviving (or not surviving) the night and the card being found (not found) are independent events. e.g P(F. GIA-P(FIA)P(GA), etc. Rob can only search one room in a given day and he has at most two days to search b. What is the probability that Rob finds the graphics card on the second day? c. What is the probability that Rob finds the graphics card? d. What is the probability that he finds the card and the card is still good?

Answer 1

The given probabilities are

P(A)1/3,P(B) 1/2,P(F|B) 2/5 2/3,P(GA) -4/5, PGB)3/5,P(F A) -

d) The required probability is,

P(Fn G)-P(FA)P (GA)P(A) + P(F|B)P(GIB)P(B) P(FnG) (1/2)(4/5)(1/3)+ (2/5)(3/5) (2/3)

c) We have using conditional independence again,

P(Fn G)-(1/2)(1-4/5)(1/3) + (2/5)(1-3/5)(23) P(Fn G) = 0.14

Using part (d) and total probability theorem,

P(F) 0.2933 P(F)-0.4333 0.14

b) We have

P(G)P(G A)P(A) +P(GB)P(B) P(G) 0.66666667

The probability,

P(FIG) = P(FnG) P(FIG)-0.6666667 P(F(G) = 8:5588667