Mathematics - Probability Question with Solution | TestHub

$$\begin{array}{lllll}{\mathrm{G}}_1=1& 4& 7& \cdots & 3\mathrm{n}-1\\ {\mathrm{G}}_2=2& 5& 8& \cdots & 3\mathrm{n}-2\\ {\mathrm{G}}_3=3& 6& 9& \cdots & 3\mathrm{n}\end{array}$$

We have $n(S)={}^{3n}{C}_2$. For $n(E)$, both numbers can be selected from the same group, or one number from ${\mathrm{G}}_2$ and one from ${\mathrm{G}}_3$.

$$\begin{array}{rl}\therefore n(E)& =3\cdot {}^n{C}_2+n{C}_1\times n{C}_1\\ & =3\cdot {}^n{C}_2+n^2\\ \therefore P(E)& =\frac{3\cdot {}^n{C}_2+n^2}{{}^{3n}{C}_2}\end{array}$$