Question

a pair of dice is thrown 6 times. if getting a total of 9 is considered as success, what is the probability of (a) 5 successes (b) atleast 5 successes

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{p}=\frac{1}{6},\mathrm{q}=1-\frac{1}{6}=\frac{5}{6},\mathrm{n}=6 \\ \mathrm{P}\left(\mathrm{X}=\mathrm{r}\right)={}^6{\mathrm{C}}_{\mathrm{r}}{\left(\frac{1}{6}\right)}^{6-\mathrm{r}}\times {\left(\frac{5}{6}\right)}^{\mathrm{r}} \\ \mathrm{r}=0,1,2,3,4,5,6 \\ \mathrm{P}\left(\mathrm{X}=5\right){=}^6{\mathrm{C}}_5{\left(\frac{1}{6}\right)}^{6-5}\times {\left(\frac{5}{6}\right)}^5={\left(\frac{5}{6}\right)}^5 \\ \mathrm{and} \\ \mathrm{P}\left(\mathrm{X}\ge 5\right)=\mathrm{P}\left(\mathrm{X}=5\right)+\mathrm{P}\left(\mathrm{X}=6\right) \\ ={\left(\frac{5}{6}\right)}^5{+}^6{\mathrm{C}}_6{\left(\frac{1}{6}\right)}^{6-6}\times {\left(\frac{5}{6}\right)}^6 \\ ={\left(\frac{5}{6}\right)}^5+{\left(\frac{5}{6}\right)}^6 \\ \mathrm{Regards} \end{gathered}$$