Solution — 2019 Question 10 Probability

$$\begin{array}{cc}& \mathrm{P}\left(\text{both red}\right)\\ & =\frac{10}{25}\left(\frac{9}{24}\right)\\ & =\frac{3}{20}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}\left(\text{different colours}\right)\\ & =1-\mathrm{P}\left(\text{same colour}\right)\\ & =1-(\frac{3}{20}+\frac{8}{25}\left(\frac{7}{24}\right)+\frac{7}{25}\left(\frac{6}{24}\right))\\ & =\frac{103}{150}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}\left(\text{at least 2 red}\right)\\ & =\mathrm{P}\left(RRR\right)+\mathrm{P}\left(RRX\right)+\mathrm{P}\left(RXR\right)+\mathrm{P}\left(XRR\right)\\ & =\frac{10}{25}\left(\frac{9}{24}\right)\left(\frac{8}{23}\right)+\frac{10}{25}\left(\frac{9}{24}\right)\left(\frac{15}{23}\right)+\frac{10}{25}\left(\frac{15}{24}\right)\left(\frac{9}{23}\right)+\frac{15}{25}\left(\frac{10}{24}\right)\left(\frac{9}{23}\right)\\ & =\frac{159}{460}∎\end{array}$$