Solution — 2024 Question 10 Probability| Binomial Distribution

$$\begin{array}{cc}& \mathrm{P}\left(\text{hits exactly once}\right)\\ & =0.7\left(0.2\right)\left(0.6\right)+0.3\left(0.4\right)\left(0.2\right)+0.3\left(0.6\right)\left(0.4\right)\\ & =0.18∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}(\text{at least 1 token}|\text{missed 2nd})\\ & =\frac{\mathrm{P}\left(\text{at least 1 token}\cap \text{missed 2nd}\right)}{\mathrm{P}\left(\text{missed 2nd}\right)}\\ & =\frac{0.7\left(0.2\right)\left(0.6\right)+0.7\left(0.2\right)\left(0.4\right)+0.3\left(0.6\right)\left(0.4\right)}{0.7\left(0.2\right)+0.3\left(0.6\right)}\\ & =0.6625\\ & =\frac{53}{80}∎\end{array}$$

$$\begin{array}{cc}& \mathrm{P}\left(6\text{ tokens}\right)\\ & =0.7\left(0.8\right)\left(0.8\right)\\ & =0.448\end{array}$$

\begin{align*} \mathrm{P}\left(Y \geq 2\right) &= 1 - \mathrm{P}\left(Y \leq 1\right) \\ &= 1 - 0.394\,25 \\ &= 0.605\,75 \\ &= 0.606\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.