Question Number 134622 by bobhans last updated on 05/Mar/21
$$\mathrm{Probability} \\ $$Initially there are 4 red balls and 3 yellow balls in a bag, however 1 ball is lost accidentally. Amy draws 2 balls from the bag at random and both of the balls drawn are red. What is the probability of the lost ball is the yellow one?
Answered by EDWIN88 last updated on 05/Mar/21
$$\mathcal{L}\mathrm{et}\:\mathrm{Y}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{event}\:\mathrm{that}\:\mathrm{lost}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{yellow}\:\mathrm{one}.\:\mathrm{P}\left(\mathrm{Y}\right)=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\mathrm{Let}\:\mathrm{Z}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{event}\:\mathrm{that}\:\mathrm{lost}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{a}\:\mathrm{red}\:\mathrm{one} \\ $$$$\mathrm{P}\left(\mathrm{Z}\right)\:=\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\mathrm{Let}\:\mathrm{X}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{event}\:\mathrm{that}\:\mathrm{the}\:\mathrm{2}\:\mathrm{balls}\:\mathrm{drawn} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{are}\:\mathrm{both}\:\mathrm{red} \\ $$$$\mathrm{The}\:\mathrm{problem}\:\mathrm{statement}\:\mathrm{requires}\:\mathrm{calculate} \\ $$$$\mathrm{P}\left(\mathrm{Y}\mid\mathrm{X}\right). \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{P}\left(\mathrm{X}\right)=\:\mathrm{P}\left(\mathrm{X}\mid\mathrm{Y}\right).\mathrm{P}\left(\mathrm{Y}\right)+\mathrm{P}\left(\mathrm{X}\mid\mathrm{Z}\right).\mathrm{P}\left(\mathrm{Z}\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{P}\left(\mathrm{X}\mid\mathrm{Y}\right)\:=\:\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{4}} /\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{6}} =\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{P}\left(\mathrm{X}\mid\mathrm{Z}\right)\:=\:\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{3}} /\mathrm{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{P}\left(\mathrm{X}\right)=\:\frac{\mathrm{2}}{\mathrm{5}}.\:\frac{\mathrm{3}}{\mathrm{7}}+\:\frac{\mathrm{1}}{\mathrm{5}}.\:\frac{\mathrm{4}}{\mathrm{7}}\:=\:\frac{\mathrm{10}}{\mathrm{35}}\:=\:\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\mathrm{note}\:\mathrm{that}\:\mathrm{P}\left(\mathrm{Y}\mid\mathrm{X}\right).\mathrm{P}\left(\mathrm{X}\right)=\:\mathrm{P}\left(\mathrm{X}\mid\mathrm{Y}\right).\mathrm{P}\left(\mathrm{Y}\right) \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{\mathrm{7}}.\mathrm{P}\left(\mathrm{Y}\mid\mathrm{X}\right)\:=\:\frac{\mathrm{2}}{\mathrm{5}}.\:\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\Leftrightarrow\:\mathrm{P}\left(\mathrm{Y}\mid\mathrm{X}\right)\:=\:\frac{\mathrm{3}}{\mathrm{5}}\: \\ $$$$ \\ $$