Question

Problem 5: Gwendolyn and Cecily are using some cards to resolve an issue: whether their absent, mutual friend Algernon is Bunburying (Cecily thinks he is). The decision protocol works as follows. Gwendolyn has 4 cards with the numbers 0, 1, 2, 3 printed on them. Cecily has 4 cards with the numbers 1, 2, 3, 4 printed on them. Gwendolyn selects three cards at random from her 4 cards, then discards the high card and the low card. The remaining card is her chosen card. For example: if 0, 2, 3 is selected, the 0 and the 3 are discarded leaving the 2 as the chosen card. Cecily does the same thing except that she discards the two highest cards. For example: if 1, 2, 3 is selected, the 2 and the 3 are discarded leaving the 1 as her chosen card. Gwendolyn and Cecily then sum the values of their chosen cards. If this sum is odd, Gwendolyn wins the dispute, otherwise Cecily wins. Part 5a: Do both Gwendolyn and Cecily has an equal chance of winning? Part 5b: If Gwendolyn wins, what is the probability that Gwendolyn's card was a 2?

Answer 1

Problem 5 7 Card Game problem on them. with nos. cecily has 4 cards with them. Gwendolyn has a cards 0, 132,3 pointed 1, 2, 3, 4 printed decision protocol, if both of them at sandom out of 4, then it Now according the chouces 3 cards can be Selected in 403 waye in each cases CASCS wile occurs For Gwendolyn, the following 023) 012) 013 ) 123. will be - For cecily, 124) Cecily, the cases 123 234. . 134, the highest She wice of Gwendolyn discards and lowest card, be left with the following card number respectively 1, 1, 2, 2. of cecily discards the two highest cards, her cases will be l- . 2 را را را 2 the can in which will be 4x4 = 16 Thus, total sample space ways be with left one card. they numbeas on for Gwendolyn to win the sum of. both of their cards should be odd (1, 2) twices (21) sia times of that will make ways, sum in this case. the sum 3, which is the only possible odd Hence total in which she can win = 8 ways. 8 Gwendolyn winning Her chances of winning of ways 2. Probability of 16 sample space This has an equal chance =%2: makes cecily's chance of winning 1-1/2 = "½, since probability of of a certain event ( here any of them winning) 1. 5a) we can see both Gwendolyn and Cecily which is Y. to win and her card to be a 56) For Gwendolyn ourselves with the ordered pair (2,1) which six times out of sixteen ways. of winning at the same time, concern shale occurs

and her card two Hence the probability of her winning being at the same time is mido