There's a contest, with 24% portuguese people participating where 50% of portugueses read a book in the last 12 months, 39% went to the cinema, and 9% into a ballet in the same period of time. We also have spanish participating with 59%, 56%, 12% respectivelly (book, cinema ,ballet).
So now the question is : What's the probability of needing less than 5 people to find 3 spanish that went to a ballet.
What I tried was the following: So what I did (with no success) was first multiply 0.76*0.12 to find the probability of going to a ballet and being spanish. Then I tried using the Binomial Distribution with 5 cases ( the 5 contestants) with x as 3 (the number of spanish people) and using the 0.75*0.12=0.09 as the probability of success, but after the calculations were all done I was still left with no correct answer. I
I also tried going with a differetn approach ( a lit bit more extense) by doing everything "manual" by making the sum of having 3 spanish people that went into a ballet in 5 contestants, then in 4 contestants and finally 3 contestants, but still with no luck.
I'm in need of a bit of light to help me solve this.
The simplest way using $bin~(4,0.09)$ is to find P(at least $3$ successes in $4$ trials), = 0.00272
Assuming that the larger example you cited is $bin(200,0.09), P(X\ge10)$, you could reduce computation using the complement, or use the normal/poisson approximation as appropriate