Find the least dart count for a strict probability target

Let n be the number of independent throws with the same success probability 0.4. Each misses with probability 0.6, so $P(\text{at least one hit in n throws})=1-0.6^n$. Follow the Teacher guidance: what happens after the first success does not affect whether at least one hit occurred.

The source’s disjoint cases are:

No condition is imposed on later throws once the first hit has occurred.

$0.4+0.6(0.4)+0.6^2(0.4)+\cdots+0.6^{n-1}(0.4)=0.4(1-0.6^n)/(1-0.6)=1-0.6^n$. Both printed approaches therefore give the same probability.

$1-0.6^n>0.9$ implies $0.6^n<0.1$, hence $n\ln0.6<\ln0.1$. Since ln 0.6 is negative, dividing reverses the sign: $n>\ln0.1/\ln0.6\approx4.5076$. The printed n>4.5 is rounded; the least integer is 5.

Evaluate $1-0.6^n$ in a calculator table, implementing the Teacher’s demonstration guidance.

The n=6 value is 0.953344 before rounding. Probability increases with n;4 fails and 5 passes. The least number of throws is 5.