<PRE><B>Question 1082068</B>
(i)  P(Drawing two cards and having at least one be a king) is the same as 1 - P(no king appears in the draw)
and the latter is easier to calculate directly:
                              P (no king in a draw of 2 cards) = (48/52)*(47/51) = 564/663
P(rawing two cards and having at least one be a king) =
                1 - 564/663 = 99/663 = {{{highlight(33/221) }}} or about 0.14932

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(ii)  Say you pick the two cards one at a time from the 4 queens:  P(red then black) + P(black then red) =
(2/4)*(2/3) + (2/4)*(2/3) = 8/12 = {{{ highlight(2/3) }}}
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This can be also be seen by the number of arrangements possible for the 4 queens:
       RRBB, RBRB, RBBR, BBRR, BRBR, BRRB  and say you just deal out the first two of these, exactly
4 out of the 6 possible arrangements will give you one R and one B.  
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