Menu Close

Question-138338




Question Number 138338 by mohammad17 last updated on 12/Apr/21
Commented by mohammad17 last updated on 12/Apr/21
help me sirlease
$${help}\:{me}\:{sirlease} \\ $$
Answered by mr W last updated on 12/Apr/21
Q1:  p=((1+3)/2^3 )=(1/2)  Q2:  p=1−(1/2^3 )=(7/8)  Q3:  p=(3/2^3 )=(3/8)
$${Q}\mathrm{1}: \\ $$$${p}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Q}\mathrm{2}: \\ $$$${p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${Q}\mathrm{3}: \\ $$$${p}=\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{8}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *