Question 1162747
You have 6 blue hats and 10 green hats.
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Probability of both hats being the same color = {{{(6C2 + 10C2)/(16C2)}}} = {{{(6!/(2!*4!)+10!/(2!*8!))/(16!/(2!*14!))}}} = {{{(15 + 45)/120}}} = {{{60/120}}} = 0.5
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If the probability of both hats being the same color is 0.5, then the probability of both hats NOT being the same color (different colors) is obviously also 0.5.
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Probability of both hats being green = {{{(10C2)/(16C2)}}} = {{{(10!/(2!*8!))/(16!/(2!*14!))}}} = {{{45/120}}} = 0.375
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Probability of one hat being blue = {{{6/16}}} = 0.375