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Question Number 111826 by Aina Samuel Temidayo last updated on 05/Sep/20

$$\mathrm{There}\mathrm{are}\mathrm{two}\mathrm{cards};\mathrm{one}\mathrm{is}\mathrm{yellow}\mathrm{on}$$$$\mathrm{both}\mathrm{sides}\mathrm{and}\mathrm{the}\mathrm{other}\mathrm{one}\mathrm{is}\mathrm{yellow}$$$$\mathrm{in}\mathrm{one}\mathrm{side}\mathrm{and}\mathrm{red}\mathrm{on}\mathrm{the}\mathrm{other}.\mathrm{The}$$$$\mathrm{cards}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{probability}\left(\frac{1}{2}\right)$$$$\mathrm{of}\mathrm{being}\mathrm{chosen},\mathrm{and}\mathrm{one}\mathrm{is}\mathrm{chosen}\mathrm{and}$$$$\mathrm{placed}\mathrm{on}\mathrm{the}\mathrm{table}.\mathrm{If}\mathrm{the}\mathrm{upper}\mathrm{side}$$$$\mathrm{of}\mathrm{the}\mathrm{card}\mathrm{on}\mathrm{the}\mathrm{table}\mathrm{is}\mathrm{yellow},$$$$\mathrm{then}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{under}$$$$\mathrm{side}\mathrm{is}\mathrm{also}\mathrm{yellow}\mathrm{is}.$$

Commented by Her_Majesty last updated on 05/Sep/20

$$case1\frac{r}{y}\left({doesn}^{\prime }tcount\right)$$$$case2\frac{y}{r}$$$$case3+4\frac{y}{y}$$$$\Rightarrow probability=\frac{2}{3}$$

Commented by Aina Samuel Temidayo last updated on 05/Sep/20

$$\mathrm{Ok}.$$

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