Mathematics - Area Under Curves Question with Solution | TestHub

Let ' $\mathrm{p}$ ' denote the probability of winning of team A whenever it plays $\therefore \mathrm{p}=\frac{2}{3}\&\mathrm{q}=1-\frac{2}{3}=\frac{1}{3}$ Let $X$ denotes the number of winning games out of 4 games i.e. $n=4$ $\therefore$ The probability of $r$ success $\mathrm{P}(\mathrm{X}=\mathrm{r})={}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}{\mathrm{p}}^{\mathrm{r}}{\mathrm{q}}^{\mathrm{n}-\mathrm{r}},\mathrm{r}=0,1,2,3,4$ $\therefore$ Probability of winning more than half games $=\mathrm{P}(\mathrm{X}>2)$ $=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)$ $={}^4{C}_3{\left(\frac{2}{3}\right)}^3\cdot {\left(\frac{1}{3}\right)}^{4-3}+{}^4{C}_4{\left(\frac{2}{3}\right)}^4\cdot {\left(\frac{1}{3}\right)}^{4-4}$ $=\frac{4!}{3!1!}\cdot \frac{8}{27}\cdot \frac{1}{3}+\frac{4!}{4!0!}\cdot \frac{16}{81}$ $=\frac{32}{81}+\frac{16}{81}=\frac{48}{81}=\frac{16}{27}$