I've read this exercise in the internet:
"The PIRON Software Company currently develops marketing software for primarily service-based organizations. They are considering expanding their operations to product marketing in the Phoenix area. The sales representatives have contacted 500 companies, and in the past 6 months, secured contracts with 75 of those companies.
Based on the scenario above, if the sales team contacted 300 more companies in the next 90 days, what is the probability of securing a contract, and why is that the probability?"
I attempted to solve this exercise but I've not done yet. If I set $X$ = "number of secured contracts in 90 days". Based on the provided assumptions, I can not find the right distribution for $X$. Does anyone has any idea or suggestion? Thank you.
It's a vague question, but here's my attempt: For every company, there's a chance $p$ that PIRON secures a contract with one company. We can assume that those 500 companies are a big enough sample, so $p$ is around $\frac{75}{500}=0.15$.
Now PIRON contacts another 300 companies, each with chance $0.15$ that they will secure a contract. As you might know, if there's a chance $p$ that an event will happen, and it is tested $n$ times, the probability that the event will happen $X$ times is: $$\binom nX \cdot p^X\cdot(1-p)^{n-X}$$ That's because of the Binomial distribution
Anyway, in our case $n=300,p=0.15,X=1$, so the chance of securing just one contract is: $$\binom{300}{1} \cdot 0.15^1\cdot(1-0.15)^{300-1}$$ As we multiply $0.85$ by itself $299$ times, the answer is tiny: according to Wolfram-Alpha it's $3.5438... × 10^{-20}$
The time doesn't seem to be of any importance.