Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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**soroban****Member**- Registered: 2007-03-09
- Posts: 451

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

soroban wrote:

*Last edited by JaneFairfax (2009-07-16 06:28:10)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

JaneFairfax wrote:

True, but that's not the main problem.

After your fix, we still have that P(B) = 2/3, so the "conclusion" still holds.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I dont see what the problem is, really (if not the minor one I pointed out). If you have two Black balls and one White ball in the bag, the probability of picking a Black ball is ⅔. If you have one Black ball and two other balls that can be either colour, the probability of picking a Black ball is also ⅔. There is nothing wrong with the probabilities of two different scenarios happening to be the same.

*Last edited by JaneFairfax (2009-07-17 02:34:53)*

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