Question Number 149981 by jlewis last updated on 08/Aug/21
$$\mathrm{a}\:\mathrm{full}\:\mathrm{deck}\:\mathrm{of}\:\mathrm{52}\:\mathrm{cards}\:\mathrm{contains}\:\mathrm{13} \\ $$$$\:\mathrm{hearts}.\:\mathrm{Pick}\:\mathrm{8}\:\mathrm{cards}\:\mathrm{from}\:\mathrm{the}\:\mathrm{deck} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{without}\:\mathrm{replacement}. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{you}\:\mathrm{get} \\ $$$$\mathrm{no}\:\mathrm{heart}? \\ $$$$ \\ $$
Answered by mr W last updated on 08/Aug/21
$${p}=\frac{{C}_{\mathrm{8}} ^{\mathrm{39}} }{{C}_{\mathrm{8}} ^{\mathrm{52}} }=\mathrm{8}.\mathrm{17\%} \\ $$
Commented by jlewis last updated on 08/Aug/21
$$\mathrm{thank}\:\mathrm{sir},\:\mathrm{and}\:\mathrm{what}\:\mathrm{of}\:\mathrm{if}\:\mathrm{its}\:\mathrm{with} \\ $$$$\:\mathrm{replacement} \\ $$
Commented by mr W last updated on 09/Aug/21
$${p}=\left(\frac{\mathrm{39}}{\mathrm{52}}\right)^{\mathrm{8}} =\mathrm{10\%} \\ $$