document.write( "Question 278057: Two cards are drawn without replacement from an ordinary deck of 52 playing cards. What are the odds against drawing a club and a diamond? \r
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document.write( "What are the odds against getting all heads or all tails in three successive flips of a coin?
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Algebra.Com's Answer #202376 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Two cards are drawn without replacement from an ordinary deck of 52 playing cards. What are the odds against drawing a club and a diamond? \n" ); document.write( "--- \n" ); document.write( "To get the odds against you need the probability for and the probability \n" ); document.write( "against. \n" ); document.write( "-- \n" ); document.write( "P(club and diamond) = (13/52)(13/51) \n" ); document.write( "P(not drawing a club and a diamond) = 1 - (13^2/(52*51)) = (52*51-13^2)/(52*51) \n" ); document.write( "- \n" ); document.write( "--- \n" ); document.write( "Then odds against = P(against)/P(for) = (52*51-13^2)/13^2 = 2483:169 \n" ); document.write( "========================================================================== \r \n" ); document.write( "\n" ); document.write( "What are the odds against getting all heads or all tails in three successive flips of a coin? \n" ); document.write( "--- \n" ); document.write( "P(all heads) = 1/8 \n" ); document.write( "P(all tails) = 1/8 \n" ); document.write( "--- \n" ); document.write( "Assuming these events are independent you get: \n" ); document.write( "P(all heads and all tails) = (1/8)^2 = 1/64 \n" ); document.write( "P(not) = 1-(1/64) = 63/64 \n" ); document.write( "--- \n" ); document.write( "Then odd against = P(not)/P(for) = 63/1 = 63:1 \n" ); document.write( "================================================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |