Question: An exhibit contains 10 items. Three of these items are randomly selected and tested, and come up positive. What are the chances that there were only three positive items and randomly selecting only those three positive items?
What I am struggling with most is the wording here.
Work so far:
(P of possible negatives): \begin{aligned}\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=0.29\end{aligned}
1- (P of possible negatives): \begin{aligned}1-0.29=0.708\end{aligned}
Thus, have I answered the question with a 70% chance, or is there more I need to consider about the chance of selecting 3 positives in a row? Any guidance/feedback would be appreciated!
P(selecting exactly those 3 items) $= 3/10*2/9*1/8=.0083333$
Alternatively you can compute it as ${3 \choose 3}/{10 \choose 3}=1/120$.
Probability of there being only 3 positive items, assuming you know that items are positive at rate p independently of one another is: ${10 \choose 3}p^3(1-p)^7$ (binomial(10,p) probability of exactly 3 successes).
So probability of both there being exactly 3 positive and you choosing them is the product of these quantities: ${10 \choose 3}p^3(1-p)^7/120=p^3(1-p)^7.$
Depending on how precise your phrasing of the question was you may have only been looking for the probability of selecting exactly 3 positive items, given only 3 of the ten items would have tested positive.
Note: ${10 \choose 3}p^3(1-p)^7$ is maximized by $p=3/10$, so we know that this quantity is less than or equal to .267, meaning $p^3(1-p)^7 <.0022236.$