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There-are-two-cards-one-is-yellow-on-both-sides-and-the-other-one-is-yellow-in-one-side-and-red-on-the-other-The-cards-have-the-same-probability-1-2-of-being-chosen-and-one-is-chosen-and-placed




Question Number 111826 by Aina Samuel Temidayo last updated on 05/Sep/20
There are two cards; one is yellow on  both sides and the other one is yellow  in one side and red on the other. The  cards have the same probability ((1/2))  of being chosen, and one is chosen and  placed on the table. If the upper side  of the card on the table is yellow,  then the probability that the under  side is also yellow is.
$$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{cards};\:\mathrm{one}\:\mathrm{is}\:\mathrm{yellow}\:\mathrm{on} \\ $$$$\mathrm{both}\:\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}\:\mathrm{is}\:\mathrm{yellow} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{side}\:\mathrm{and}\:\mathrm{red}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other}.\:\mathrm{The} \\ $$$$\mathrm{cards}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{probability}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{of}\:\mathrm{being}\:\mathrm{chosen},\:\mathrm{and}\:\mathrm{one}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{and} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{side} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{card}\:\mathrm{on}\:\mathrm{the}\:\mathrm{table}\:\mathrm{is}\:\mathrm{yellow}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{under} \\ $$$$\mathrm{side}\:\mathrm{is}\:\mathrm{also}\:\mathrm{yellow}\:\mathrm{is}. \\ $$
Commented by Her_Majesty last updated on 05/Sep/20
case 1 (r/y) (doesn′t count)  case 2 (y/r)  case 3+4 (y/y)  ⇒ probability=(2/3)
$${case}\:\mathrm{1}\:\frac{{r}}{{y}}\:\left({doesn}'{t}\:{count}\right) \\ $$$${case}\:\mathrm{2}\:\frac{{y}}{{r}} \\ $$$${case}\:\mathrm{3}+\mathrm{4}\:\frac{{y}}{{y}} \\ $$$$\Rightarrow\:{probability}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by Aina Samuel Temidayo last updated on 05/Sep/20
Ok.
$$\mathrm{Ok}. \\ $$

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