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Vishnu Nanilal Panicker replied to BMAD's question in New Logic/Math PuzzlesIf the number of apples you'd have to eat were the same from both the bowls, of course anyone would have chosen from the one with the less number of poisonous apples. But here's the case is different. Let's analyse: 1st bowl We can use simple probability. Death happens when both apples you choose are poisonous. If any of the two apples were non poisonous, you'd survive. P(First apple to be poisonous) = 3/5. P(Second apple to be poisonous given first apple was poisonous)= (3/5)*(2/4). =30 % So, there's a 30 percentage chance that you'd die, eating from the1st bowl. 2nd bowl This is more complicated, since there are 3 tries which can result in 2 possibilities, viz. poisonous or non poisonous. Assume, Poisonous is 1 and Non poisonous is 0. So the combinations are: 000,001,010,011 100,101,110,111 Here 111 is not possible, since there are only 2 poisonous apples in bowl 2. So there are 7 possible outcomes. Out of these, only the ones with two 1s can cause death. They are: 011,101,110. So effectivey 3/7. Around 42%. So I'd choose the one with more number of poisonous apples.