Question 1168391
this is 6C3*0.4^3*0.6^3=0.2765

this is 4 or more would be 1-(3 or fewer): binomcdf(10,.4,3)=0.3822
0.2150 for 3
0.1209 for 2
0.0403 for 1
0.006 for 0
that is 0.3822 

at most (not almost) 7 would be everything fewer than or equal to 6. That is 0.9452.

Normal approximation
np=100*0.4=40 mean
np(1-p)=40*0.6=24=variance
SD = sqrt (V)=4.90
at least 55: z>(55-40)/4.9 or >3.06. that has probability of 0.0011
exact is 0.0017 (1-binomcdf(100,0.4,54))ENTER