Every year for Halloween, it is estimated that $5$ out of every $8$ houses give away candy.
(a) If we want to receive candy from $5$ different houses, what is the probability that we will need to visit at least $10$ of them?
I tried using a negative binomial distribution approach.
We are allow to use R to solve the problem.
Hence, I have $negbinom(5,5/8)$
I let X be the # of houses until we receive candy from 5
To find $P(X>=10)$ I use the cdf $1-P(X<=9)$
Using R I calculated $[1-pnbinom(4,5,5/8)]=0.2166$
Is any of this correct?
The idea is right. I have not checked the numerical computation.
Another way to solve the problem is to note that the probability we need to visit at least $10$ houses is the probability we get $4$ or fewer successes (candy) in the first $9$ houses we visit. This is a binomial distribution problem, with answer $$\sum_{k=0}^4 \binom{9}{k}(5/8)^k(3/8)^{9-k}.$$