Question 1130264
I find for this type of problem it is easier to count the number of ways the event(s) of interest
can occur then divide by the total number of possible outcomes.<br>

The number of selections of 4 numbers from the set (0, 1, . . ., 12), with replacement, is {{{ 13^4 }}}.  This will be the denominator in our probability calculation.<br>

Note: Using notation (nCr  = n!/((n-r)!r!))<br>

a) There are {{{ 4!*13C4 }}} ways to pick 4 unique numbers from the set of 13 (4! because A,B,C,D
and C,B,D,A are the same pick but picked in different order).   

P(no matches) = {{{ 4!*13C4 / 13^4 = 17160 / 28561 = highlight( 1320 / 2197 ) }}} or about {{{ highlight( 0.6008 ) }}}

<br>

b) P(no more than two match) = P(no matches) + P(exactly two match) = 1320/2197 + P(exactly two match)

P(exactly two match) = {{{ (13C3)(4C2)*2! / 13^4 = 3432 / 28561 = 264 / 2197 }}}  or about <b>0.12016</b><br>


P(no more than two match) = {{{ (1320+264) / 2197  = highlight( 1584 / 2197 ) }}} or about {{{ highlight( 0.7210 ) }}} <br>


This is probably the answer your teacher is looking for. <br>

I'd like to point out: if {A,B,C,D} are any four unique numbers selected from the set {0,1, . . ., 12} then part (b) addresses A,B,C,D (and its permutations) and A,A,B,C  (and its permutations) while the pattern A,A,B,B (and its permutations) are NOT counted.  For example, {7,7,11,11} is NOT counted in part (b) even though technically there are only two matching numbers.  When I interpret "not more than 2 are the same" I sometimes think {7,7,11,11} should be included.  <br>

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EDITED:  Fixed the wording on the paragraph about number of selections of 4 numbers, and added this section:<br> 

To include the pattern {A,A,B,B}, add to the above answer: 
  P(AABB) = {{{ (13C2)(4C2) / 13^4 = 468 / 28561 }}}  to part (b).