Question

a pair of dice is thrown 7 times. if getting a total of 7 what is the probability of (1)exactly 6 success(2)at most 6 success

Answer 1

$$\begin{gathered} \mathrm{Sample}\mathrm{space}=\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right)....\left(1,6\right),\left(2,1\right),\left(2,2\right)...\left(2,6\right),\left(3,1\right)....\left(3,6\right)....\left(6,6\right)\right\} \\ \mathrm{probability}\mathrm{of}\mathrm{getting}7\mathrm{as}\mathrm{a}\mathrm{sum}\mathrm{happens}\mathrm{in}\left\{\left(1,6\right),\left(6,1\right),\left(2,5\right),\left(5,2\right),\left(3,4\right),\left(4,3\right)\right\}. \\ \mathrm{Thus},\mathrm{probability}\mathrm{of}\mathrm{getting}7\mathrm{as}\mathrm{total}\mathrm{is}\frac{6}{36}=\frac{1}{6} \\ \mathrm{i})\mathrm{Probability}\mathrm{of}\mathrm{exactly}6\mathrm{successes}\mathrm{out}\mathrm{of}7={}^7\mathrm{C}_6{\left(\frac{1}{6}\right)}^6\left(\frac{5}{6}\right)=\frac{35}{6^7} \\ \mathrm{ii})\mathrm{Probability}\mathrm{of}\mathrm{atmost}6\mathrm{successes}=1-{}^7\mathrm{C}_7{\left(\frac{1}{6}\right)}^7{\left(\frac{5}{6}\right)}^0=1-\frac{1}{6^7} \end{gathered}$$