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P(both boys | 1 boy) = P(1 boy | both boys) * P(both boys) / P(1 boy)
P(1 boy | both boys) = 1 as %100 possibility P(both boys) = 1/2 * 1/2 as 0.5 probability for each child without prior information P(1 boy) = P(1 boy | both boys) * P(both boys) + P(1 boy | NOT[both boys]) * P(NOT[both boys]) = 1 * 1/4 + 2/3 * 3/4 = 3/4 Therefore, P(both boys | 1 boy) = (1 * 1/4) / 3/4 = 1/3 All possibilities: AP = [g,b; b,g; b,b; g,g];
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Day 3: Conditional Probability
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P(both boys | 1 boy) = P(1 boy | both boys) * P(both boys) / P(1 boy)
P(1 boy | both boys) = 1 as %100 possibility P(both boys) = 1/2 * 1/2 as 0.5 probability for each child without prior information P(1 boy) = P(1 boy | both boys) * P(both boys) + P(1 boy | NOT[both boys]) * P(NOT[both boys]) = 1 * 1/4 + 2/3 * 3/4 = 3/4 Therefore, P(both boys | 1 boy) = (1 * 1/4) / 3/4 = 1/3
All possibilities: AP = [g,b; b,g; b,b; g,g];