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Question Number 145831 by nadovic last updated on 08/Jul/21

$$\mathrm{A}\mathrm{box}\mathit{P},\mathrm{contains}4\mathrm{white},2\mathrm{green}\mathrm{and}$$$$3\mathrm{blue}\mathrm{cards}.\mathrm{Another}\mathrm{box}\mathit{Q},\mathrm{contains}$$$$2\mathrm{white},3\mathrm{green}\mathrm{and}2\mathrm{blue}\mathrm{cards}.\mathrm{A}\mathrm{card}$$$$\mathrm{is}\mathrm{picked}\mathrm{at}\mathrm{random}\mathrm{from}\mathit{P}\mathrm{and}\mathrm{placed}$$$$\mathrm{in}\mathit{Q}.\mathrm{A}\mathrm{card}\mathrm{is}\mathrm{then}\mathrm{picked}\mathrm{from}\mathit{Q}.$$$$\mathrm{Find}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}$$$$\left(a\right)\mathrm{card}\mathrm{picked}\mathrm{from}\mathit{Q}\mathrm{is}\mathrm{white}.$$$$\left(b\right)\mathrm{cards}\mathrm{picked}\mathrm{from}\mathit{P}\mathrm{and}\mathit{Q}\mathrm{are}\mathrm{of}$$$$\mathrm{the}\mathrm{same}\mathrm{colour}.$$

Answered by bramlexs22 last updated on 08/Jul/21

$$\left(\mathrm{a}\right)\mathrm{P}\left(\mathrm{A}\right)=\frac{4}{9}\times \frac{3}{8}+\frac{5}{9}\times \frac{2}{8}=\frac{22}{72}$$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{11}{36}$$

Answered by bramlexs22 last updated on 08/Jul/21

$$\left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{A}\right)=\frac{4}{9}\times \frac{3}{8}+\frac{2}{9}\times \frac{4}{8}+\frac{3}{9}\times \frac{3}{8}$$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{12+8+9}{72}=\frac{29}{72}$$

Commented by nadovic last updated on 09/Jul/21

$$ThankyouSir$$

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