document.write( "Question 724790:  If two cards are drawn without replacement from an ordinary deck, find the probabilities of the following results.\r
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document.write( "Q. The second is a heart, given that the first is a heart.\r
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document.write( "This question asks me to solve for \r
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document.write( "P(Second is a heart given that the first is a heart)= P(E and F)/ P(F)\r
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document.write( "I tried solving the first part \"The second is a heart\" 
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document.write( "by doing this way. Let E: second is a heart
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document.write( "P(E) = 13/52.\r
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document.write( "At this stage, i'm completely lost how to get the second part \" given that the first is a heart\" \r
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document.write( "In this case, what should i do? 
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| Algebra.Com's Answer #443910 by stanbon(75887)      You can put this solution on YOUR website! If two cards are drawn without replacement from an ordinary deck, find the probabilities of the following results. \n" ); document.write( "Q. The second is a heart, given that the first is a heart. \n" ); document.write( "This question asks me to solve for \n" ); document.write( "P(Second is a heart given that the first is a heart)= P(E and F)/ P(F) \n" ); document.write( "I tried solving the first part \"The second is a heart\" \n" ); document.write( "by doing this way. Let E: second is a heart \n" ); document.write( "P(E) = 13/52. \n" ); document.write( "At this stage, i'm completely lost how to get the second part \" given that the first is a heart\" \n" ); document.write( "----- \n" ); document.write( "P(heart AND heart) = 13C2/52C2 = 1/17 \n" ); document.write( "--- \n" ); document.write( "P(heart|heart) = (1/17)/(13/52) = (1/17)(52/13) = 0.2353 \n" ); document.write( "-------------------------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "============= \n" ); document.write( " \n" ); document.write( " |