Question

There are two identical cabinets, A and B, each with five drawers. Cabinet A has one empty drawer, two drawers with a gold coin in each and two drawers with a silver coin in each. Cabinet B has two empty drawers, two drawers with a silver coin in each and one drawer with a gold coin in it. You and your friend are given a task of finding a gold coin by opening a maximum of three drawers. Suppose that your friend chooses a cabinet at random and opens two of its drawers at random. You find that one drawer is empty and that the other drawer contains a silver coin. To maximise your chances of finding a gold coin in the final atempt, would you open another drawer of the same cabinet or would you choose to open a drawer from the other cabinet? If you follow your strategy, what is the probabilty of finding a gold coin by opening one more drawer? If you follow the alternative strategy to yours, what is then the probabilty of finding a gold coin by opening one more drawer?

Answer 1

B I WIGI LG LS JLS J ULLS LL LGJ. Probability of choosing gold con from dicines A given that a silver and empty diawer is opened is 8/3 Since we are left with 3 diawass of which & have gold coins and one has silver coin. Probability of choosing a gold can giren that are empty and one silver con has been opened from B is 1/3 since we are we are left with one gold, one Silwer, one empty desiwas doonis

same choose cabinet to Suppose the strategy used by me is to sme cabinet from which my biend has doawn two drawess Thin event of finding gold coin chawen is a = (GOA) o Ca - P(GIA) Plat P(GOA) ☆ PC G/B) P(13) = PlanB). Hina P(a) = P(GOA) + P (GNB) = PCGLAD PCA) +PC61B) PCB) - / * /2 + 1/ x / 2 + 1% = %='la = 0

Suppose the strategy of chawing drawer from cabinet is used G = P(GRA) TP(GOB) Gnother then, = P(Gla) P(A) TP(G/13) PC B) Suppose that a cabinet is dawn by my friend, then if I disaw another cabinet B the probability of choosing gold is 1/5. Similaaly if drawes from cabinet B 3 diawn by forend, I choose cabinst A. Then probability of getting gold con ic 2/5 PCG) A) = 1/- & P(G) 13 ) = 2/;- henee. P(G)- PCG/A) P(A) + P(GIB) P(B) - 1/x % + / 12 1 + 2 = $3 = 0.3 - To 10 10

Since P(a) in (1) is greate than P(a) in 2, I choose to choose strategy of choosing same cabinet. If others alternative in choosen I have probability Chasing old loin to be 0.3.