4 ms·

I remember this one from a youth mathematics book. 10⁻⁴ chance of one bomb, and 10⁻⁸ of two (assuming the other one is an independent event — which it would be,

by gattr 4y ago

I remember this one from a youth mathematics book. 10⁻⁴ chance of one bomb, and 10⁻⁸ of two (assuming the other one is an independent event — which it would be, as you're certainly not in collusion with any actual bomber).

dismalpedigree 4y ago
Potentially catastrophic false assumption made at the end there.
meindnoch 4y ago
Conditional probability doesn’t work like that. If you always have a bomb with you, then the probability of you sitting on a plane with a bomb is 1, and thus the probability of you sitting on a plane with another bomb is again 10^-4.
- 542458 4y ago
  Yes, the post you’re replying to was recounting a joke :)
  - gattr 4y ago
    Indeed. Hmm, but what if I roll a D20 before each flight and take a bomb only if I get a "1"? Do I decrease the average two-bombs probability to 5·10⁻⁶?
    - meindnoch 4y ago
      Yes. But the conditional probability of the plane having a bomb that’s not yours will still be 10^-4.
- deleted 4y ago
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