An insurance salesperson knows that the more contacts they make with potential customers, the greater the opportunity to sell a policy. If the probability that a person buys an insurance policy after the visit is constant and equal to 0.25, and if the set of visits constitutes an independent set of trials, how many potential buyers must the salesperson visit so that the probability of selling at least one policy is 0.80?
You should visit 6, I thought of it as follows:
the probability of selling at least one policy to 6 people is $1-$ the probability of not selling any policy. It is $1-0.75^6=0.82$
Let random variable $X$ represent the number of customers, such that:
$$Pr(X=x)=\binom{n}{x} \times p^x \times (1-p)^{n-x}$$
This situation is a binomial distribution. In this case, we are required to solve for $n$ given that we know $p=0.25$ and $Pr(X \ge 1) \ge 0.8$:
$$Pr(X \ge 1)=1-Pr(X=0)$$ So, if we know that $Pr(X \ge 1) \ge0.8$, then $Pr(X=0)\le 0.2$
Now, we can solve $Pr(X=0)\le 0.2$ for $n$
$$0.2 \ge\binom{n}{0} \times 0.25^0 \times (0.75)^{n}$$
$$0.2 \ge (0.75)^{n}$$
$$n \ge \log_{0.75}{0.2}$$
$$n \ge 5.5945$$
So, as the number of customers is only defined for integers, the salesman must have $6$ customers