document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193236>Question 1193236</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (a) Let X be a random variable giving the number of aces in a random draw of 4 cards from an ordinary deck 52 cards. Construct a table showing the probability distribution of X. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #825251 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3816)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=825251\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "There are four aces in a deck. One of each suit (hearts, diamonds, clubs, spades).\r<BR>\n" );
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document.write( "X = number of aces in a four card hand\r<BR>\n" );
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document.write( "The allowed values of X are drawn from the set {0,1,2,3,4}\r<BR>\n" );
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document.write( "If X = 0, then we have 52-4 = 48 non-ace cards to pick from.<BR>\n" );
document.write( "We have four slots to fill. Order does not matter in card hands, meaning we use the nCr combination formula.<BR>\n" );
document.write( "Plug in n = 48 and r = 4<BR>\n" );
document.write( "n C r = (n!)/(r!(n-r)!)<BR>\n" );
document.write( "48 C 4 = (48!)/(4!*(48-4)!)<BR>\n" );
document.write( "48 C 4 = (48!)/(4!*44!)<BR>\n" );
document.write( "48 C 4 = (48*47*46*45*44!)/(4!*44!)<BR>\n" );
document.write( "48 C 4 = (48*47*46*45)/(4!)<BR>\n" );
document.write( "48 C 4 = (48*47*46*45)/(4*3*2*1)<BR>\n" );
document.write( "48 C 4 = (4669920)/(24)<BR>\n" );
document.write( "48 C 4 = 194580<BR>\n" );
document.write( "There are 194,580 ways to select four non-ace cards\r<BR>\n" );
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document.write( "Next, calculate the number of ways to select four cards regardless if we get an ace or not.<BR>\n" );
document.write( "n = 52 and r = 4<BR>\n" );
document.write( "n C r = (n!)/(r!(n-r)!)<BR>\n" );
document.write( "52 C 4 = (52!)/(4!*(52-4)!)<BR>\n" );
document.write( "52 C 4 = (52!)/(4!*48!)<BR>\n" );
document.write( "52 C 4 = (52*51*50*49*48!)/(4!*48!)<BR>\n" );
document.write( "52 C 4 = (52*51*50*49)/(4!)<BR>\n" );
document.write( "52 C 4 = (52*51*50*49)/(4*3*2*1)<BR>\n" );
document.write( "52 C 4 = (6497400)/(24)<BR>\n" );
document.write( "52 C 4 = 270725<BR>\n" );
document.write( "We have 270,725 ways to select any four cards whether we get an ace or not.\r<BR>\n" );
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document.write( "I should mention that the cards are NOT put back. We consider this scenario as \"no replacement\".\r<BR>\n" );
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document.write( "Once we have the two results calculated earlier, we divide them<BR>\n" );
document.write( "194580/270725 = 0.718737\r<BR>\n" );
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document.write( "The probability that we get zero aces, i.e. X = 0, is roughly 0.718737\r<BR>\n" );
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document.write( "In the table below, we'll have P(X) = 0.718737 pair up with X = 0 in the first row.<BR>\n" );
document.write( "There's roughly a 71.8737% chance of getting no aces in a four-card hand.\r<BR>\n" );
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document.write( "Now onto X = 1 which means we have exactly one ace in our four-card hand.\r<BR>\n" );
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document.write( "We have four aces, so there are four ways to select exactly one of them.<BR>\n" );
document.write( "You can use the nCr formula to note that 4C1 = 4, though this may be a bit overkill.\r<BR>\n" );
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document.write( "For the remaining r = 4-1 = 3 slots, we have n = 52-4 = 48 cards to pick from<BR>\n" );
document.write( "n C r = (n!)/(r!(n-r)!)<BR>\n" );
document.write( "48 C 3 = (48!)/(3!*(48-3)!)<BR>\n" );
document.write( "48 C 3 = (48!)/(3!*45!)<BR>\n" );
document.write( "48 C 3 = (48*47*46*45!)/(3!*45!)<BR>\n" );
document.write( "48 C 3 = (48*47*46)/(3!)<BR>\n" );
document.write( "48 C 3 = (48*47*46)/(3*2*1)<BR>\n" );
document.write( "48 C 3 = (103776)/(6)<BR>\n" );
document.write( "48 C 3 = 17296<BR>\n" );
document.write( "which is the number of ways to pick the other 3 non-ace cards.\r<BR>\n" );
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document.write( "We found there are 4 ways to pick the single ace, and 17296 ways to pick the other three cards.<BR>\n" );
document.write( "This then means there are 4*17296 = 69184 ways to get a four-card hand such that we have exactly one ace.\r<BR>\n" );
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document.write( "Divide this over the 270725 figure found in the previous section (it represents the number of four-card hands total)<BR>\n" );
document.write( "69184/270725 = 0.255551<BR>\n" );
document.write( "This is the approximate P(X) value for X = 1<BR>\n" );
document.write( "Refer to the second row of the table below.\r<BR>\n" );
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document.write( "Next up is X = 2<BR>\n" );
document.write( "We have 4C2 = 6 ways to select the two aces and 48C2 = 1128 ways to pick the other two non-ace cards.<BR>\n" );
document.write( "I'm skipping steps, but you'd use the nCr formula as described in the other sections above (with different n and r values of course).\r<BR>\n" );
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document.write( "So we have (4C2)*(48C2) = (6)*(1128) = 6768 ways to select exactly two aces and two non-aces.<BR>\n" );
document.write( "This leads to 6768/270725 = 0.025000 as the approximate P(X) value when X = 2<BR>\n" );
document.write( "The more accurate P(X) value is P(X) = 0.02499953827684 but I'm rounding results to six decimal places. <BR>\n" );
document.write( "Feel free to use different decimal precision. Of course, if your teacher instructs otherwise, then use that precision.\r<BR>\n" );
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document.write( "For X = 3, we have 4C3 = 4 ways to pick the three aces. <BR>\n" );
document.write( "This is the same as having 4 ways to not pick a particular ace.<BR>\n" );
document.write( "Then we have 48C1 = 48 ways to pick the non-ace.\r<BR>\n" );
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document.write( "There are (4C3)*(48C1) = (4)*(48) = 192 ways to pick exactly three aces and some other card.<BR>\n" );
document.write( "192/270725 = 0.000709 approximately<BR>\n" );
document.write( "Therefore, if X = 3 then P(X) = 0.000709 roughly\r<BR>\n" );
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document.write( "Lastly, we need to find P(X) when X = 4<BR>\n" );
document.write( "There's only one way to pick four aces. You could use the fact that 4C4 = 1.\r<BR>\n" );
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document.write( "This is out of 270725 ways to select a four-card hand<BR>\n" );
document.write( "1/270725 = 0.000004 approximately<BR>\n" );
document.write( "The more accurate value is 0.0000036937852 but I'm rounding to 6 decimal places like mentioned earlier.\r<BR>\n" );
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document.write( "You probably were able to spot the pattern how to calculate any general P(X) value fairly quickly.\r<BR>\n" );
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document.write( "That pattern would be:<BR>\n" );
document.write( "P(X) = ( (4 C w)*(48 C v) )/(52 C 4)<BR>\n" );
document.write( "where w+v = 4 and w,v are integers from the set {0,1,2,3,4}<BR>\n" );
document.write( "w = number of aces, v = number of non-aces\r<BR>\n" );
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document.write( "For example, we could have w = 0 and v = 4<BR>\n" );
document.write( "P(X) = ( (4 C w)*(48 C v) )/(52 C 4)<BR>\n" );
document.write( "P(X) = ( (4 C 0)*(48 C 4) )/(52 C 4)<BR>\n" );
document.write( "P(X) = ( (1)*(194,580) )/(270,725)<BR>\n" );
document.write( "P(X) = 0.71873672545941<BR>\n" );
document.write( "P(X) = 0.718737<BR>\n" );
document.write( "which matches up with the P(X) value we found when X = 0\r<BR>\n" );
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document.write( "<font color=red>Answer:</font><TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >X</TD><TD >P(X)</TD></tr><tr><TD >0</TD><TD >0.718737</TD></tr><tr><TD >1</TD><TD >0.255551</TD></tr><tr><TD >2</TD><TD >0.025000</TD></tr><tr><TD >3</TD><TD >0.000709</TD></tr><tr><TD >4</TD><TD >0.000004</TD></tr>\n" );
document.write( "</TABLE>where X is the number of aces of a four-card hand\r<BR>\n" );
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document.write( "The P(X) values should add up to 1, or very close to it. <BR>\n" );
document.write( "In this case, <BR>\n" );
document.write( "0.718737+0.255551+0.025000+0.000709+0.000004 = 1.000001<BR>\n" );
document.write( "which unfortunately is due to rounding error; however, we're close enough.<BR>\n" );
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