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windsylph

  • 3 years ago

Probability: Joe is a fool with probability 0.6, a thief with probability 0.7, and neither with probability 0.25. Determine the probability that he is a fool or a thief but not both.

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  1. apple_pi
    • 3 years ago
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    What is probability of Joe being not a thief. Similarly what is probability of not being a fool?

  2. windsylph
    • 3 years ago
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    My answer is 0.2, since the probability of Joe being a thief, a fool, or both a thief and a fool is 1-0.25 = 0.75, and 0.75 = P(Joe is thief) + P(Joe is fool) - P(Joe is thief and fool) = 0.6 + 0.7 - x. x = 0.55, and (0.6 + 0.7) - 0.55 = 0.2, which is the probability of Joe being a thief or a fool but not both. Please confirm?

  3. windsylph
    • 3 years ago
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    wait, I meant 0.75 - 0.55, not 0.6 + 0.7 - 0.55

  4. windsylph
    • 3 years ago
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    because 0.75 is already thief, fool, or both, and 0.55 is only both, so if i subtract 0.55 from 0.75, I get thief OR fool, right?

  5. apple_pi
    • 3 years ago
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    Ok first off, when we 'add' probabilities so like P(THIS and THAT) we multiply P(THIS) and P(THAT) The way I would do it is get P(fool but not thief) + P(thief but not fool)

  6. windsylph
    • 3 years ago
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    Sorry, for not being proper, but how would I get P(fool but not thief) and P(thief but not fool) if the event that Joe is a thief and Joe is a fool overlap?

  7. apple_pi
    • 3 years ago
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    OK to get P(fool not thief) get P(fool) = 0.6, P(not thief) = 1- P(thief) = 0.3 now multiply the two to get P(fool not thief). Do similarly for the other one. Then just add the two probabilities. Sorry but I have to go now.

  8. windsylph
    • 3 years ago
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    It's okay, thank you for your help.

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