Algebra.Com's Answer #784601 by ikleyn(52943)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
.
\n" );
document.write( "You take the four Aces, four 2's, and four 3's from a standard deck of 52 cards, forming a set of 12 cards.
\n" );
document.write( "You then deal all cards at random to four players, so that each player gets three cards.
\n" );
document.write( "What is the probability that each player gets an Ace, a 2, and a 3?
\n" );
document.write( "~~~~~~~~~~~~~\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " Thank you for your comment. Under its influence, I reconsidered and redone my previous solution.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " This corrected and re-edited solution is placed below.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " Under the previous location\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1161100.html\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " I also updated my post, so now you see the corrected (and identical) version here and there.\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "\r\n" );
document.write( "The total number of ways to split 12 cards in triples (12 = 3 + 3 + 3 + 3) is\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " T =  . . . = . . . =  = 369600.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The favorable triples are those that have NO repeating.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "More precisely, each favorable splitting is the set of 4 (four) triples, such that no one has repeating.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now our task is to calculate the total number of such favorable splittings.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "In favorable splitting, 1-st triple may contain any one of four Aces, any one of four 2's and any one of four 3's;\r\n" );
document.write( " so, 1-st triple can be constructed by 4*4*4 =  ways.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " 2-nd triple may contain any one of the remaining three Aces, any one of the remaining three 2's \r\n" );
document.write( " and any one of the remaining three 3's; so, 2-nd triple can be constructed by 3*3*3 =  ways.\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( " next, 3-rd triple may contain any one of remaining two Aces, any one of remaining two 2's \r\n" );
document.write( " and any one of remaining two 3's; so, 3-rd triple can be constructed by 2*2*2 =  ways.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " finally, 4-th triple may exist only one: there is no any choice for it.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Hence, the number of all different favorable splittings is  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Therefore, the probability under the problem's question is\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( " P =  =  =  = 0.037403. ANSWER\r\n" );
document.write( " \r
\n" );
document.write( "\n" );
document.write( "Solved.\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( " |