document.write( "Question 1201887: A drawer contains 5 red socks and three black socks. A sock is taken out at random and not replaced. A second sock is then taken out. Draw a tree diagram and calculate the probability that either pair of red socks or a pair of black socks is chosen.
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Algebra.Com's Answer #836439 by ikleyn(52782) $\"\"$ $\"About$

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\n" ); document.write( "A drawer contains 5 red socks and three black socks. A sock is taken out at random and not replaced.
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document.write( "We want to calculate the probability having a pair of red socks or a pair of black socks\r\n" );
document.write( "P = P(RR or BB).\r\n" );
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document.write( "First of all, it is clear that these events RR or BB are disjoint: they have empty intersection,\r\n" );
document.write( "which means that they can not happen simultaneously.\r\n" );
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document.write( "Therefore, P(RR or BB) = P(RR) + P(BB) :  the probability of this combined event is the sum \r\n" );
document.write( "of probabilities of particular events.\r\n" );
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document.write( "So, we can calculate two probabilities P(RR) and P(BB) separately and then add them.\r\n" );
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document.write( "          Let's calculate P(RR).\r\n" );
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document.write( "Originally, we have 5 red socks among 5+3 = 8 socs. So the probability to draw 1st sock red is  .\r\n" );
document.write( "After that, we have 4 red socks and 8-1=7 socks remaining. So, the probability to draw 2nd sock red is  .\r\n" );
document.write( "The probability to have both socks read after two drawing is the product   = .\r\n" );
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document.write( "          Let's calculate P(BB).  The logic is very similar.\r\n" );
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document.write( "Originally, we have 3 black socks among 5+3 = 8 socs. So the probability to draw 1st sock black is  .\r\n" );
document.write( "After that, we have 2 black socks and 3-1 = 2 socks remaining. So, the probability to draw 2nd sock black is  .\r\n" );
document.write( "The probability to have both socks black after two drawing is the product   = .\r\n" );
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document.write( "Finally, we shoud add the two found probabilities. We get then\r\n" );
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document.write( "    P(RR or BB) =  +  =  = .    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "There are other ways to solve this problem, but this way is the simplest:
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