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Question Number 128103 by byaw last updated on 04/Jan/21

$$$$$$\mathrm{A}\mathrm{box}\mathrm{contains}4\mathrm{black}5\mathrm{white}$$$$\mathrm{and}6\mathrm{red}\mathrm{shirts}.3\mathrm{are}\mathrm{drawn}\mathrm{from}$$$$\mathrm{the}\mathrm{box}\mathrm{one}\mathrm{after}\mathrm{the}\mathrm{other}$$$$\mathrm{without}\mathrm{replacement}.\mathrm{Find}\mathrm{the}$$$$\mathrm{proability}\mathrm{that}\mathrm{two}\mathrm{of}\mathrm{the}\mathrm{same}$$$$\mathrm{color}\mathrm{and}\mathrm{one}\mathrm{a}\mathrm{different}\mathrm{color}.$$

Answered by liberty last updated on 04/Jan/21

$$\mathrm{A}=\{(2\mathrm{b},1\mathrm{w}),(2\mathrm{b},1\mathrm{r}),(2\mathrm{r},1\mathrm{w}),(2\mathrm{r},1\mathrm{b}),(2\mathrm{w},1\mathrm{b}),(2\mathrm{w},1\mathrm{r})\}$$$$\mathrm{p}\left(\mathrm{A}\right)=3\times (\frac{4}{15}.\frac{3}{14}(\frac{5}{13}+\frac{6}{13})+\frac{6}{15}.\frac{5}{14}(\frac{5}{13}+\frac{4}{13})+\frac{5}{15}.\frac{4}{14}(\frac{4}{13}+\frac{6}{13}))$$$$=3\times (\frac{12.11}{15.14.13}+\frac{30.9}{15.14.13}+\frac{20.10}{15.14.13})$$$$=\frac{132+270+200}{14.13.5}=\frac{602}{910}=\frac{301}{455}=\frac{43}{65}$$$$$$

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