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Old 07-26-2007   #5
Daniel F
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Tougher than it looks. Here we go:Flip a coin. You get H (heads) or T (tails). So two possible outcomes in one flip. If you flip it 5 times, you have 2^5=32 possible outcomes.How many of these 32 outcomes contain exactly 3 heads?When we have three heads, we must also have exactly three tails, so your goal is to determine how many combinations of this there are. Consider one option:HHHTTFirst we need to flip three heads in a row. The odds of this happening are (1/2)^3=1/8Then we need to flip two tails in a row. Odds: (1/2)^2=1/4. So the probability of getting HHHTT is (1/8)(1/4)=1/32 (good general idea to remember in probability problems: x=and, +=or).Now consider another option: HHTHT. We flip 2 heads (odds: (1/2)^2=1/4), 1 tails (odds: 1/2), 1 heads (1/2), and another tails (1/2). So the odds of getting this combination are (1/4)(1/2)(1/2)(1/2)=1/32.The point is, the probability of getting each combination of 3 H's and 2 T's is always the same, 1/32. So, how many possible 5 flip combinations have 3 heads? Well, that would be solved as a combination, 5C3=5!/(3!2!)=10. So the answer is 10(1/32)=10/32=5/16
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