document.write( "Question 344260: Suppose you have 6 coins in your pocket, 3 pennies and 3 nickels. You reach in and take out 3 of
\n" ); document.write( " the six coins at random; then you throw those 3 onto a table. Calculate the probability that at the
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Algebra.Com's Answer #246294 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let's use this notation for each possibility PXY and  NXY, where\r\n" );
document.write( "the first letter tells whether it's a penny or a nickel. The X will be\r\n" );
document.write( "either 1, 2, or 3 to indicate which of the the three pennies or which of the\r\n" );
document.write( "three nickels it is.  The Y will represent either heads H or tails T.  \r\n" );
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document.write( "For example:\r\n" );
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document.write( "P2H means penny #2 comes up heads, that is X=2 and Y=H\r\n" );
document.write( "and\r\n" );
document.write( "N3T means nickel #3 comes up tails, that is, X=3 and Y=T\r\n" );
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document.write( "First we calculate the numerator of the probability.\r\n" );
document.write( "So we count the number of successful ways we can have \r\n" );
document.write( "2 pennies and 1 nickel, then 1 penny and 2 nickels, and \r\n" );
document.write( "then 3 nickels and no pennies. \r\n" );
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document.write( "1.  2 pennies and 1 nickel\r\n" );
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document.write( "PXY PXY NXY\r\n" );
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document.write( "We can choose the X's for the pennies in 3C2 or 3 ways\r\n" );
document.write( "We can then assign one of the pennies Y=H or Y=T in 2 ways\r\n" );
document.write( "We can then assign the other penny Y=H or Y=T in 2 ways \r\n" );
document.write( "We can then choose the X for the nickel in 3 ways\r\n" );
document.write( "We can then choose the Y for the nickel in only 1 way (heads) Y=H\r\n" );
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document.write( "That's 3*2*2*3*1 = 36 ways\r\n" );
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document.write( "2. 1 penny and 2 nickels.\r\n" );
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document.write( "PXY  NXY  NXY\r\n" );
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document.write( "We can choose the X for the penny in 3 ways\r\n" );
document.write( "We can then assign the penny Y=H or Y=T in 2 ways \r\n" );
document.write( "We can choose X's for the 2 nickels in 3C2 or 3 ways\r\n" );
document.write( "We can choose the Y's for the nickels in 3 ways, that is, give one of them\r\n" );
document.write( "T and the other H in 2 ways, and give them both H's which is 1 more way. \r\n" );
document.write( "That's 3 ways.  \r\n" );
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document.write( "That's 3*2*3*3 = 54\r\n" );
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document.write( "3. 3 nickels and no pennies.\r\n" );
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document.write( "N1Y  N2Y  N3Y\r\n" );
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document.write( "We'll first figure the number of ways to make the Y's anything and then\r\n" );
document.write( "subtract the 1 case N1T N2T N3T of all tails.\r\n" );
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document.write( "There are 2 ways to choose the Y for N1.\r\n" );
document.write( "Then there are 2 ways to choose Y for N2.\r\n" );
document.write( "Then there are 2 ways to choose Y for N3.\r\n" );
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document.write( "That's 2*2*2 or 8 ways minus the 1 way in which all 3 are tails.\r\n" );
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document.write( "So that's 7 ways to have 3 nickels\r\n" );
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document.write( "So the total number of successful ways is 36+54+7=97\r\n" );
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document.write( "Now we calculate the denominator of the probability.\r\n" );
document.write( "So we count the number of possible ways we can have \r\n" );
document.write( "3 pennies no nickels, then 2 pennies and 1 nickel, then 1 penny \r\n" );
document.write( "and 2 nickels, and 3 nickels and no pennies. \r\n" );
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document.write( "1.  3 pennies and no nickels.\r\n" );
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document.write( "P1Y  P2Y  P3Y\r\n" );
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document.write( "We can assign the first penny Y=H or Y=T, that's 2 ways.\r\n" );
document.write( "We can then assign the second penny Y=H or Y=T, that's 2 ways\r\n" );
document.write( "We can then assign the third penny Y=H or Y=T, that's 2 ways.\r\n" );
document.write( "So the number of ways we can have 3 pennies is 2*2*2=2³=8 ways.\r\n" );
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document.write( "2.  2 pennies and 1 nickel\r\n" );
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document.write( "PXY PXY NXY\r\n" );
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document.write( "We can choose the X's for the pennies in 3C2 or 3 ways\r\n" );
document.write( "We can then assign one of the pennies Y=H or Y=T in 2 ways\r\n" );
document.write( "We can then assign the other penny Y=H or Y=T in 2 ways \r\n" );
document.write( "We can then choose the X for the nickel in 3 ways\r\n" );
document.write( "We can then choose the Y for the nickel Y=H or Y=T in 2 ways\r\n" );
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document.write( "That's 3*2*2*3*2 = 72 ways\r\n" );
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document.write( "3. 1 penny and 2 nickels.  This is esactly like the previous\r\n" );
document.write( "case, just swapping nickes and pennies.  But I'll go through\r\n" );
document.write( "it anyway, but it'll come out the same, 72.\r\n" );
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document.write( "PXY  NXY  NXY\r\n" );
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document.write( "We can choose the X for the penny in 3 ways\r\n" );
document.write( "We can then assign the penny Y=H or Y=T in 2 ways \r\n" );
document.write( "We can choose X's for the 2 nickels in 3C2 or 3 ways\r\n" );
document.write( "We can then assign one of the nickels Y=H or Y=T in 2 ways\r\n" );
document.write( "We can then assign the other nickel Y=H or Y=T in 2 ways \r\n" );
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document.write( "That's 3*2*3*2*2 = 72\r\n" );
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document.write( "4. 3 nickels.  This is the same as for when there are 3 pennies above,\r\n" );
document.write( "but I'll go through it anyway:\r\n" );
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document.write( "N1T  N2T  N3T\r\n" );
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document.write( "We can assign the first nickel Y=H or Y=T, that's 2 ways.\r\n" );
document.write( "We can then assign the second nickel Y=H or Y=T, that's 2 ways\r\n" );
document.write( "We can then assign the third nickel Y=H or Y=T, that's 2 ways.\r\n" );
document.write( "So the number of ways we can have 3 nickels is 2*2*2=2³=8 ways.\r\n" );
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document.write( "So the total number of possible ways is 8+72+72+8 = 160\r\n" );
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document.write( "So the desired probability is 97/160\r\n" );
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document.write( "Edwin
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