DivyaD wrote:A fair die has sides labeled with 1, 2, 3, 4, 5, and 6 dots. If the die is rolled 4 times, what is the probability that on at least one roll, the die will show a 6?

$$ a) \frac{1}{6}\,\,\,\,\,\,\, b) \frac{625}{1296}\,\,\,\,\,\,\,c) \frac{649}{1296}\,\,\,\,\,\,\,d) \frac{671}{1296}\,\,\,\,\,\,\,e) \frac{2}{3}$$

$$? = 1 - P\left( {\underbrace {{\rm{no}}\,\,{\rm{6}}\,\,{\rm{in}}\,\,{\rm{all}}\,\,{\rm{4}}\,\,{\rm{times}}}_{{\rm{unfavorable}}}} \right)$$
$$\left. \matrix{
{\rm{Total}}\,\,:\,\,\,{6^4}\,\,{\rm{equiprobable}}\,\,{\rm{outcomes}}\,\,\, \hfill \cr
{\rm{unfav}}\,\,{\rm{:}}\,\,{5^4}\,\,{\rm{of}}\,\,{\rm{them}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,P\left( {{\rm{unfavorable}}} \right) = {\left( {{5 \over 6}} \right)^4}$$
$$? = {{{6^4} - {5^4}} \over {{6^4}}} = {{\left( {{6^2} + {5^2}} \right)\left( {6 + 5} \right)\left( {6 - 5} \right)} \over {{6^4}}} = {{61 \cdot 11} \over {{6^4}}} = {{610 + 61} \over {{6^4}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.