Question 919688
<pre>
(a)
For him to roll 2 6's, he will 

EITHER

1. choose a weighted die    <--probability = 2/3
AND
2. roll a 6 the first time  <--probability = 1/7
AND
3. roll a 6 the second time <--probability = 1/7

OR

1. choose a non-weighted die  <--probability = 1/3
AND
2. roll a 6 the first time  <--probability = 1/6
AND
3. roll a 6 the second time <--probability = 1/6

AND suggests multiplication and OR suggests addition:

{{{(2/3)(1/7)(1/7)+(1/3)(1/6)(1/6)}}}{{{""=""}}}{{{121/5292}}}

Approximately 0.023 

---------
 (b)

                                  P(weighted and (6 and 6) )
P(weighted die | rolled 2 6's) = ------------------------
                                      P(6 and 6)

The denominator is the result of part (a) 

= {{{((2/3)(1/7)(1/7))/(121/5292)}}} = {{{(2/147)/(121/5292)}}} = {{{72/121}}} = about 0.595

Edwin</pre>