document.write( "Question 4020: Find the probability of drawing an ace and a jack when two cards are drawn (without replacement) from a standard deck of cards \n" ); document.write( "
Algebra.Com's Answer #1778 by longjonsilver(2297)\"\" \"About 
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you don't say if either order is OK or an ace then a jack. I will do both:\r
\n" ); document.write( "\n" ); document.write( "1. ORDER matters:
\n" ); document.write( "P(ace THEN jack) = P(ace) + P(jack)\r
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\n" ); document.write( "\n" ); document.write( "so, first, there are 4 aces and 52 cards ---> P(ace) = 4/52 = 1/13.
\n" ); document.write( "then there are 4 jacks and 51 cards --> P(jack) = 4/51\r
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\n" ); document.write( "\n" ); document.write( "so, P(ace then jack) = 1/13 + 4/51 = 0.155 to 3dp\r
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\n" ); document.write( "\n" ); document.write( "How about P(jack then ace)?..the same numbers apply, so P=0.155 too.\r
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\n" ); document.write( "\n" ); document.write( "2. ORDER does not matter:
\n" ); document.write( "Now P(ace and a jack in any order) = P(ace then jack) or P(jack then ace), which is 2 times 0.155 --> 0.311 (use the full number in the calculation, not 0.155, as this is an approximate value).\r
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\n" ); document.write( "\n" ); document.write( "jon. \n" ); document.write( "
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