Using binomial distrubution,
p=0.30
![P(X=x) = C(n,x)p^x (1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%20%3D%20C%28n%2Cx%29p%5Ex%20%281-p%29%5E%7Bn-x%7D)
where
![C(n,x)=\frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C%28n%2Cx%29%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
(1) exactly x=5 customers out of n=6 make purchase
![P(X=x)=C(n,x)p^x (1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%3DC%28n%2Cx%29p%5Ex%20%281-p%29%5E%7Bn-x%7D)
![P(X=5)=C(6,5)(0.3)^5 (0.7)^{6-5}](https://tex.z-dn.net/?f=P%28X%3D5%29%3DC%286%2C5%29%280.3%29%5E5%20%280.7%29%5E%7B6-5%7D)
![=6*0.00243*0.7](https://tex.z-dn.net/?f=%3D6%2A0.00243%2A0.7)
![=0.10206](https://tex.z-dn.net/?f=%3D0.10206)
(2) at least three customers make a purchase
![P(X\ge 3)=\sum_{i=3}^{6}C(6,i)(0.3)^i (1-0.3)^{6-i}](https://tex.z-dn.net/?f=P%28X%5Cge%203%29%3D%5Csum_%7Bi%3D3%7D%5E%7B6%7DC%286%2Ci%29%280.3%29%5Ei%20%281-0.3%29%5E%7B6-i%7D)
=0.185220+0.059535+0.010206+0.000729=0.255690
(3) Two or less customers make purchase
Similarly, using binomial distribution
![P(X\le2)=\sum_{i=3}^{6}C(6,i)(0.3)^i(1-0.3)^{6-i}](https://tex.z-dn.net/?f=P%28X%5Cle2%29%3D%5Csum_%7Bi%3D3%7D%5E%7B6%7DC%286%2Ci%29%280.3%29%5Ei%281-0.3%29%5E%7B6-i%7D)
=0.117649+0.302526+0.324135
=0.744310
In fact, we do not need to use the binomial distribution to find the answer.
From results in (2), we have
P(X<=2)=1-P(X>=3)=1-0.255690=0.744310
(4) At least one customer makes a purchase
Use the same formulas above for P(X>=1)
Here we will just use
P(X>=1)
= 1- P(X<1)
= 1-P(X=0)
=1-0.117649
=0.882351