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Taciav

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  1. Probabilities of something occuring in sequence multiply together, they don't add. We could calculate the probability where we don't eat 2 poisonous or where we do. I chose to calculate where we do. For the first bowl. We only care if you eat 2 poisonous apples. The probability the first apple eaten is poisonous is 3/5. Assuming you ate a poisonous apple because that is the only scenario we care about, the probability the second apple is poisonous is 2/4. This is the only scenario eating from this bowl that we die. Multiply this together (3/5 * 2/4) and the chance of dying is 30%. 70% chance you survive. For the second bowl, we do the same thing, but there are multiple possibility branches for this one. The probability the first apple eaten is poisonous is 2/5. The probability the second apple is poisonous is 1/4 if you already ate a poisonous one. So a 10% chance you die after eating the second apple (assuming it is fast poison). If you are still alive then the 2nd apple was not poisonous so that had a probability of 3/4. That makes he probability for the third equal to 1/3. So 2/5 * 3/4 * 1/3 is 10% chance of dying. The chance the first apple is not poisonous is 3/5. So for the second the chance is 2/4, leaving the last at 1/3. So multiply these out and you get 10%%. Due to having 3 scenarios that result in death we have to add those probabilities. 10% + 10% + 10% = 30%. 70% chance we survive. So both bowls have the same odds. Others arrived at this answer, but with the wrong math. Cheers.
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