I can’t get my head around the correct way of beginning to solve this question:
It is known that $25\%$ of 11-year-old children have no decayed, missing, or filled (DMF) teeth. In a sample of $10$ children, find the probability that exactly $3$
(a) have DMF teeth.
(b) do not have DMF teeth.
I know that $\Pr(DMF) = .75$ and $\Pr(DMF*)=.25$, but I am confused about the sample part. I don’t have a strategy to do it.
Answers are: $0.003$, $0.25$
Thanks in advance!
If we assume that each child is equally likely to have decayed, missing, or filled teeth, we can treat this as a binomial distribution problem with $\Pr(DMF) = 0.75$.
The binomial distribution states that the probability of obtaining exactly $k$ successes in $n$ trials, each of which has probability $p$ of being successful, is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $\binom{n}{k}$ counts the number of ways exactly $k$ successes can occur in $n$ trials, $p^k$ is the probability of $k$ successes, and $(1 - p)^k$ is the probability of $n - k$ failures.
For the probability of exactly three children having DMF teeth, $p = 0.75$ and $k = 3$.
For the probability of exactly three children having no DMF teeth, $p = 1 - \Pr(DMF) = 1 - 0.75 = 0.25$ and $k = 3$.