BTGmoderatorDC wrote:A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
Source: Magoosh
$$?\,\, = \,\,P\left( {{\rm{same}}\,\,{\rm{ball}}\,\,{\rm{twice}}\,{\rm{,}}\,{\rm{with}}\,\,{\rm{replacement}}} \right)$$
Exactly 1 ball extracted in the second extraction (out of 10 equiprobable possibilities) will be the same as (any given) ball extracted in the first extraction.
Therefore the answer is 1/10, immediately.
Another approach:
$${\rm{Total}} = 10 \cdot 10\,\,{\rm{pairs}}\,\,\,\left( {{1^{{\rm{st}}}}\,,\,\,{2^{{\rm{nd}}}}} \right)\,{\rm{ extracted}}\,,\,\,{\rm{all}}\,\,{\rm{equiprobable}}$$
$${\rm{Favorable}}\,\,{\rm{ = }}\,\,{\rm{10}} \cdot {\rm{1}}\,\,{\rm{pairs}}\,{\rm{among}}\,{\rm{the}}\,{\rm{above}}\,\,\,\,\,\left[ {\,\left( {{1^{{\rm{st}}}} = {\rm{any}}\,\,{\rm{ball}}\,,\,\,{2^{{\rm{nd}}}} = {\rm{same}}\,\,{1^{{\rm{st}}}}\,\,{\rm{ball}}} \right)\,} \right]\,$$
$$? = {{10} \over {100}} = {1 \over {10}}$$
These solutions follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
