a pair of fair dice is tossed 3 times. find the probability a seven appears 3 times  

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Virginia - First let's look at what goes in the denominator ... this is called your "sample space". The first dice has 6 possible outcomes, agree?  They are {1,2,3,4,5,6}. The second dice also has the same 6 possible outcomes. Since the tossing of the first and second dice are "independent" events, you simply multiply to get the total number of outcomes ( 6 ways times 6 ways = 36 possible outcomes).  Try writing out the 36 possible outcomes to convince yourself this makes sense. Finally, tossing the "pair" of dice 3 times results in 36x36 x36 possible outcomes so your numerator (the total sample space) will equal 46,656 possible outcomes. Next, let's look at the numerator.  Well, we have {(1,6),(2,5),(3,4),(6,1),(5,2),(4,3)}, where the first value is the number of dots on the first dice and the second value is the number of dots on the second dice.  In total, we have 6 possible outcomes that add up to 7 on one toss or 6x6x6 = 216 ways when tossing the dice 3 times. Hope this helps!