Probability - Catching a fish

Question

I'm not sure I'm understanding this question,

If the probability of me to catch a fish is 0.8, what is the minimal tries must I make so that the probability of catching a fish on at least once is at least 0.9?

I thought maybe I need to do here some kind of probability tree? that catching fish and me catching fish is -> 0.8*0.9 = 0.72, and at least one time means to use the binomial distribution? I'm not sure cause I don't have a limit to choose my options from here.

Thanks!

Answer 1

Hint. Try to use this property $$P(\text{x to happen})=1-P(\text{x not to happen})$$

Thus, if the probability of catching a fish in one try is $p$ then

• the probability of not catching a fish in one try is $1-p$
• the probability of not catching a fish in $2$ tries is $(1-p)^2$
• ...
• the probability of not catching a fish in $n$ tries is $(1-p)^n$

Then the probability to catch at least one fish in $n$ tries is $$1-(1-p)^n$$ Now, the question is to find minimal $n$ such that $$1-(1-0.8)^n\geq 0.9$$