Solutions - 5060-02

(a)

$$\begin{array}{c}P[\text{1st correct}\cap \text{2nd incorrect}]=\frac{6}{10}\cdot \frac{4}{9}\\ \\ \gg \gg \frac{24}{90}\approx 0.2667\end{array}$$

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(b)

$$\begin{array}{c}\left(\genfrac{}{}{0px}{}{6}{4}\right)=\frac{6!}{4!2!}\gg \gg 15\end{array}$$

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(c)

$$\begin{array}{c}P[\text{4 errors in first 6}]=\frac{\left(\genfrac{}{}{0px}{}{6}{4}\right)}{\left(\genfrac{}{}{0px}{}{10}{4}\right)}\\ \\ \gg \gg \frac{15}{210}\gg \gg \approx 0.0714\end{array}$$