document.write( "Question 392066: two cards are drawn from a 52card deck without replacement. what is the probability that the 2nd card is a heart? \n" ); document.write( "

Algebra.Com's Answer #278260 by Edwin McCravy(20086) $\"\"$ $\"About$

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two cards are drawn from a 52card deck without replacement. what is the probability that the 2nd card is a heart?
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document.write( "The easy way is to see that it's just one out of four suits or .\r\n" );
document.write( "That's because no matter what is done with the cards before drawing a card, as\r\n" );
document.write( "long as the results of what's done with the cards before drawing that second\r\n" );
document.write( "card is not given, then for any one card to come up a heart is still one out of\r\n" );
document.write( "four suits.\r\n" );
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document.write( "However some people find that hard to grasp.  So we can do it the long way.\r\n" );
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document.write( "There are 52 cards, 13 of which are hearts and 39 of which are non-hearts.\r\n" );
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document.write( "Let's describe the entire event, which is:\r\n" );
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document.write( "(1st a non-heart AND 2nd card a heart) OR (1st card a heart AND 2nd card a heart)\r\n" );
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document.write( "We want:\r\n" );
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document.write( "P[(1st a non-heart AND 2nd card a heart) OR (1st card a heart AND 2nd card a heart)]  \r\n" );
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document.write( "Remembering that AND implies \"multiply\" and OR implies \"add\", we see that the\r\n" );
document.write( "above equals:\r\n" );
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document.write( "P(1st a non-heart)*P(2nd card a heart) + P(1st card a heart)*P(2nd card a heart)\r\n" );
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document.write( "So it comes out the same the long way as when we immediately write the answer\r\n" );
document.write( "as one suit out of 4 from the start.\r\n" );
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document.write( "Edwin

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