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Question Number 117754 by bemath last updated on 13/Oct/20

Commented by bemath last updated on 13/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by bobhans last updated on 13/Oct/20

$$\left(\mathrm{a}\right)\mathrm{P}(\mathrm{B},\mathrm{B})+\mathrm{P}(\mathrm{R},\mathrm{R})+\mathrm{P}(\mathrm{Y},\mathrm{Y})={\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2=\frac{1}{3}$$$$\left(\mathrm{b}\right)\mathrm{P}(\mathrm{B},\mathrm{R})+\mathrm{P}(\mathrm{R},\mathrm{R})+\mathrm{P}(\mathrm{Y},\mathrm{R})+\mathrm{P}(\mathrm{R},\mathrm{B})+\mathrm{P}(\mathrm{R},\mathrm{Y})=$$$$5\times {\left(\frac{1}{3}\right)}^2=\frac{5}{9}.$$

Answered by MJS_new last updated on 13/Oct/20

$$\mathrm{same}\mathrm{colour}:\mathrm{first}\mathrm{draw}\mathrm{is}\mathrm{any}\mathrm{colour},$$$$\mathrm{chance}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{of}\mathrm{bag}2\mathrm{is}\frac{1}{3}$$$$\mathrm{at}\mathrm{least}1\mathrm{red}:1-\left(\mathrm{no}\mathrm{red}\right)=1-\frac{2}{3}\times \frac{2}{3}=\frac{5}{9}$$

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