document.write( "Question 379323: A newspaper advertises six different concerts, four different plays, and two different basketball games. If a couple selects three activities at random, find the probability that they will attened:\r
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Algebra.Com's Answer #269526 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "12 different choices of which 6 are different concerts P(a concert) = 6/12 = 1/2
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document.write( "P(attending three concerts) = 6/12 * 5/11 * 4/10 \r
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document.write( "12 different choices of which 4 are different plays and 2 are different BB games
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document.write( "P(Two plays and one basketball game) = 4/12 * 3/11 * 2/10\r
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document.write( "P(One movie, one play, and one basketball game) = 0   (movie not a option) \n" );
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