Question 328413
a) P(none)={{{(0.2)^6=0.000064}}}
.
.
.
b)At least one means "not none".
P(not none)+P(none)={{{1}}}
P(not none)=1-P(none)={{{1-0.000064=0.999936}}}
.
.
.
c) Find P(3 or greater successes)={{{P(3)+P(4)+P(5)+P(6)}}}
{{{P(0)=1*(0.8)^0*(0.2)^6=0.000064}}}
{{{P(1)=6*(0.8)^1*(0.2)^5=0.001536}}}
{{{P(2)=15*(0.8)^2*(0.2)^4=0.01536}}}
{{{P(3)=20*(0.8)^3*(0.2)^3=0.08192}}}
{{{P(4)=15*(0.8)^4*(0.2)^2=0.24576}}}
{{{P(5)=6*(0.8)^5*(0.2)^1=0.39322}}}
{{{P(6)=1*(0.8)^6*(0.2)^0=0.26214}}}
P(>=3)={{{P(3)+P(4)+P(5)+P(6)}}}
P(>=3)={{{0.08192+0.24576+0.39322+0.26214=highlight(0.983)}}}
Since {{{P>0.9}}}, success.
.
.
.
We could have also calculated,
{{{P(0)+P(1)+P(2)=0.000064+0.001536+0.01536=0.01696}}}
and subtracted from 1
{{{P=1-P(0)-P(1)-P(2)=1-0.01696=0.983}}}
Same answer but saving one step.