Derive a cookie success rate and compare two minimum-packet methods

Let $X\sim B(8,p/100)$. The mean condition $8p/100=3.2$ gives $p=40$, so $X\sim B(8,0.4)$. Then $q=P(X\leq4)=0.8263296$, giving $0.826$ (3 s.f.).

Let $Y\sim B(12,q)$ count packets with at most four winning slips. More than ten packets means $Y\geq11$. Then $P(Y>10)=P(Y=11)+P(Y=12)\approx0.3569729192=0.357$ (3 s.f.). The source intermediate approximation is .3569729193; the final rounded answer agrees.

Here the packet event changes to exactly four slips: $r=P(X=4)=0.2322432$. For $W\sim B(n,r)$, require $P(W\geq1)>0.9$, equivalently $(1-r)^n<0.1$. Hence $n>\ln(0.1)/\ln(0.7677568)\approx8.7125979$, giving least $n=9$.

The source checks the zero-success probability directly:

The zero-success probability decreases as n increases. Eight fails and nine passes, so the least n is 9, agreeing with the logarithmic method.