document.write( "Question 425117: Two cards are drawn, with replacement, from a standard 52-card deck. Find the expected number of diamonds \n" ); document.write( "
Algebra.Com's Answer #296142 by Edwin McCravy(20086) $\"\"$ $\"About$
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document.write( "The other tutor's answer is incorrect.\r\n" );
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document.write( "The possibilities are \r\n" );
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document.write( "1. 0 diamonds, that is, a non-diamond the first time and \r\n" );
document.write( "   a non-diamond the second time.\r\n" );
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document.write( "2. 1 diamond, that is, a diamond the first time and a non-diamond \r\n" );
document.write( "   the second time, or a non-diamond the first time and a diamond \r\n" );
document.write( "   the second time.\r\n" );
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document.write( "3. 2 diamonds, that is, a diamond the first time and a diamond\r\n" );
document.write( "   the second time.\r\n" );
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document.write( "Remember that \"AND\" usually indicates that you are to multiply \r\n" );
document.write( "probabilities, while \"OR\" indicates that you are to add them.\r\n" );
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document.write( "The probability of drawing a diamond is 1/4 and the probability of \r\n" );
document.write( "drawing a non-diamond is 3/4\r\n" );
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document.write( "1. Find the probability of 0 diamonds, (two non-diamonds)\r\n" );
document.write( "   P(non-diamond first AND non-diamond second) =\r\n" );
document.write( "   P(non-diamond first)*P(non-diamond second)\r\n" );
document.write( "   (3/4)(3/4) = 9/16 \r\n" );
document.write( "   \r\n" );
document.write( "2. Find the probability of exactly 1 diamond (and 1 non-diamond)\r\n" );
document.write( "   P[(diamond first AND non-diamond second) OR (non-diamond first\r\n" );
document.write( "   AND diamond second) =\r\n" );
document.write( "   P(diamond first)*P(non-diamond second) + \r\n" );
document.write( "                P(non-diamond first)*P(diamond second) =\r\n" );
document.write( "   (1/4)(3/4) + (3/4)(1/4) = 3/16 + 3/16 = 6/16 = 3/8 \r\n" );
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document.write( "3. Find the probability of 2 diamonds:\r\n" );
document.write( "   P(diamond first AND diamond second) =\r\n" );
document.write( "   P(diamond first)*P(diamond second) =\r\n" );
document.write( "   (1/4)(1/4) = 1/16\r\n" );
document.write( "   \r\n" );
document.write( "Next we make a discrete probability distribution function:\r\n" );
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document.write( "Number of   \r\n" );
document.write( "diamonds    Probability\r\n" );
document.write( "   x           P(x)\r\n" );
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document.write( "   0           9/16\r\n" );
document.write( "   1            3/8 \r\n" );
document.write( "   2           1/16\r\n" );
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document.write( "                 1   \r\n" );
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document.write( "Notice that the sum of the probabilities of all possible events is\r\n" );
document.write( "9/16 + 3/8 + 1/16 = 9/16 + 6/16 + 1/16 = 16/16 = 1,\r\n" );
document.write( "which you should always check. \r\n" );
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document.write( "To find the expectation of the variable x:\r\n" );
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document.write( "E(x) = ∑[x*P(x)] = 0*(9/16) + 1*(3/8) + 2*(1/16) = \r\n" );
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document.write( "0 + 3/8 + 2/16 = 3/8 + 1/8 = 4/8 = 1/2\r\n" );
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document.write( "So the expectation is 1/2.\r\n" );
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document.write( "[Note: \"Expectation\" means what you would expect to average if\r\n" );
document.write( "you repeated the experiment every day for a long period of time\r\n" );
document.write( "and took the average of the number of diamonds you drew. It does\r\n" );
document.write( "NOT mean that you would ever \"expect\" to get half a diamond. \r\n" );
document.write( "However if you drew two cards without replacement every day for\r\n" );
document.write( "365 days, you might get say 181 diamonds, and you would have\r\n" );
document.write( "averaged 181/365 or .49589 diamonds per day, which is very near .5\r\n" );
document.write( "or 1/2.]\r\n" );
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document.write( "Edwin
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