Question Number 220740 by Spillover last updated on 18/May/25
Answered by mr W last updated on 18/May/25
$$\mathrm{52}\:{cards}:\:\mathrm{26}\:{red}\:{and}\:\mathrm{26}\:{black} \\ $$$${to}\:{select}\:\mathrm{13}\:{cards}\:{there}\:{are}\:{totally} \\ $$$${C}_{\mathrm{13}} ^{\mathrm{52}} \:{possibilities} \\ $$$${to}\:{select}\:\mathrm{13}\:{red}\:{cards}\:{there}\:{are}\:{C}_{\mathrm{13}} ^{\mathrm{26}} \\ $$$${possibilities} \\ $$$$\Rightarrow{p}=\frac{{C}_{\mathrm{13}} ^{\mathrm{26}} }{{C}_{\mathrm{13}} ^{\mathrm{52}} }=\frac{\mathrm{19}}{\mathrm{1}\:\mathrm{160}\:\mathrm{054}}\approx\frac{\mathrm{1}}{\mathrm{61055}} \\ $$