Question 837396
A box contains 3 red marbles and 6 blue marbles and 1 white marble.
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I will guess at something you may have wanted that you didn't tell us.
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A. TWO marbles are selected at random, and the first one is not replaced
   before the second one is drawn. Find the probability:

10. P(Blue 1st then Red 2nd) = {{{(6/11)(3/10)}}} = {{{18/110}}} = {{{9/55}}}

11. P(Blue 1st then Blue 2nd) = {{{(6/11)(5/10)}}} = {{{30/110}}} = {{{3/11}}}
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B. THREE marbles are selected at random,, one at a time, and are not 
   replaced.  Find the probability:

12. P(Red 1st, White 2nd, then Blue 3rd) = {{{(3/11)(1/10)(6/9)}}} = {{{18/990}}} = {{{1/55}}} 

13. P(Red 1st, Red 2nd then Red 3rd) = {{{(3/11)(2/10)(1/9)}}} = {{{6/990}}} = {{{1/165}}}

14. P(White 1st, Red 2nd, then White 3rd) = 0 
    That's because there is only 1 white marble. Since we don't replace the
    first white marble, it is impossible to get it third.

Edwin</pre>