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There are 4 red and 6 blue marbles in a bag, 3 are picked out at random, what is the probability that atleast one of them is red? My answer is 1/2 First I calculated the total number of ways that I can pick out three marbles at random: 10C3 = 120. Then I multiplied: 4C1 * 6C2 = 60, I simplified 60/120 and got 1/2, can someone please confirm if I am correct or not, thank you in advance, I got this question off of an exam paper and I don't have an answer sheet for it, thank you.

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Your answer would have been correct if the question was 'find the probability that EXACTLY one of them is red'. But it says (if you wrote it correctly) that you must determine the probability that AT LEAST one is red. So this can be done by inverting the probability that NONE of them are red: $$ P(\mbox{none are red})=\frac{6C3}{10C3}=\frac{20}{120}=\frac{1}{6} $$ so then you get $$ P(\mbox{at least one is red})=1-\frac{1}{6}=\frac{5}{6} $$

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  • $\begingroup$ Thanks so much! Makes much more sense! $\endgroup$ – S.E. Chahine Dec 5 '13 at 10:52

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