In a bag there are 3 Snicker's bars and 2 Milky Way bars. What is the probability of choosing at least one Snickers when choosing two consecutive candy bars randomly?\nOptions:\n(A) 9/10\n(B) 16/20\n(C) 2/5\n(D) 3/5\n(E) ½ Hmmm, my stream of consciousness:
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P(at least one Snicker's Bar) = 1 - P(no Snicker's Bar, so 2 Milky Way Bars) = 1- 2/5*1/4 = 9/10. So, the answer is (A).
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