Previous in Permutation and Combination Next in Permutation and Combination
Question Number 98027 by bemath last updated on 11/Jun/20
$$\mathrm{One}\:\mathrm{card}\:\mathrm{is}\:\mathrm{randomly}\:\mathrm{selected} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{pack}\:\mathrm{of}\:\mathrm{52}\:\mathrm{playing}\:\mathrm{cards}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{picture}\:\mathrm{card}.\: \\ $$
Commented by bobhans last updated on 11/Jun/20
$$\mathrm{Let}\:\mathrm{P}_{\mathrm{i}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{event}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{a}\:'\mathrm{picture}\:\mathrm{card}' \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{3}\:\mathrm{picture}\:\mathrm{cards}\:\mathrm{in}\:\mathrm{each}\:\mathrm{suit}, \\ $$$$\mathrm{so}\:\mathrm{n}\left(\mathrm{P}_{\mathrm{i}} \right)\:=\:\mathrm{12}\:\therefore\:\mathrm{P}\left(\mathrm{P}_{\mathrm{i}} \right)\:=\:\frac{\mathrm{12}}{\mathrm{52}}\:=\frac{\mathrm{3}}{\mathrm{13}} \\ $$