I got some help with this type of problem and am trying to do one on my own now, can someone look this over really good and see if im doing this correctly please. im sorry, i don't know how to add powers and all that so i just typed it 
APPROXIMATLY 10.3% OF AMERICAN HIGH SCHOOL STUDENTS DROP OUT OF SCHOOL BEFORE GRADUATION. CHOOSE 10 STUDENTS ENTERING HIGH SCHOOL AT RANDOM. FIND THE PROBABILITY THAT AT LEAST 6 GRADUATE.
n=10
p=.103
q=.897
x= 6,7,8,9,10
p(6)= 10!/(10-6)!6! (.6)to the 6th power (.4) to the fourth power=.251
p(7)= 10!/(10-7)!7! (.6)to the seventh power (.4) to the third power = .215
p(8)= 10!/(10-8)!8! (.6) to the eighth power (.4) squared= .121
p(9)= 10!/(10-9)!9! (.6) to the ninth power (.4) to the first power= .040
p(10)= 10!/(10-10)!10! (.6) to the tenth power (.4) to the 0 power= .006
(.251)(.215)(.121)(.006)= .633
one little mistake and it messes EVERYTHING up...am i on the right track? thak you
APPROXIMATLY 10.3% OF AMERICAN HIGH SCHOOL STUDENTS DROP OUT OF SCHOOL BEFORE GRADUATION. CHOOSE 10 STUDENTS ENTERING HIGH SCHOOL AT RANDOM. FIND THE PROBABILITY THAT AT LEAST 6 GRADUATE.
n=10
p=.103
q=.897
x= 6,7,8,9,10
p(6)= 10!/(10-6)!6! (.6)to the 6th power (.4) to the fourth power=.251
p(7)= 10!/(10-7)!7! (.6)to the seventh power (.4) to the third power = .215
p(8)= 10!/(10-8)!8! (.6) to the eighth power (.4) squared= .121
p(9)= 10!/(10-9)!9! (.6) to the ninth power (.4) to the first power= .040
p(10)= 10!/(10-10)!10! (.6) to the tenth power (.4) to the 0 power= .006
(.251)(.215)(.121)(.006)= .633
one little mistake and it messes EVERYTHING up...am i on the right track? thak you
