What is the probability that for any $9$ randomly chosen apples, $3$ of the apples will be rejected? STATEMENT: At a food processing plant, the best apples are bagged to be sold in grocery stores. The remaining apples are either thrown out if damaged or used in food products if not appealing enough to be bagged and sold. If apples are randomly chosen for a special inspection, the probability is $0.2342$ that the $3rd$ rejected apple will be the $9th$ apple randomly chosen.
QUESTION: What is the probability that for any $9$ randomly chosen apples, $3$ of the apples will be rejected?
MY WORKING:
If we let rejected apple to be $success$ then data suggests that $r = 3rd$ $Successs$ to occur on $x = 9th$ trial has probability $0.2342$, which is problem involving Negative Binomial Distribution. By using its probability distribution we have: $f(x)={x-1 \choose r-1}p^r(1-p)^{x-r}$
By putting values of $r=3$, $x=9, f(x=9)=0.2342$, and solving for $p$ we get its value. From here on, I am not sure how to tackle the question to answer required result.
Any guidance will be appreciated. Thanks
 A: The question is actually impossible, but here is a solution which ignores that:

*

*you have said the negative binomial probability that  the $3$rd rejected apple will be the $9$th apple randomly chosen is ${x-1 \choose r-1}p^r(1-p)^{x-r}$


*similarly the binomial probability that that for any $9$ randomly chosen apples $3$ of the apples will be rejected is ${x \choose r}p^r(1-p)^{x-r}$


*dividing the latter by the former gives $\frac xr=3$ so the answer should be $3\times 0.2342 = 0.7026$
But this is all nonsense:  $p^r(1-p)^{x-r}$ and any positive constant multiple of it is maximised with $p \in [0,1]$ when $p=\frac rx$ which is $\frac13$ here, so the negative binomial probability cannot exceed $28\frac{2^6}{3^9} \approx 0.09105$ contrary to the the question and the binomial probability cannot exceed $84\frac{2^6}{3^9} \approx 0.27313$ contrary to the answer.
A: Let $p$ is probability that randomly chosen apple will be rejected, $P(k,n)$ is probability that among first $k$ randomly chosen $k$ apples will be exactly $n$ rejected, $Q(k,n)$ is probability that $n$-th rejected apple will be $k$-th randomly chosen apple.
$$P(k,n)=P(k-1,n-1)p+P(k-1,n)(1-p), P(0,0)=1, P(0,m>0)=0$$
Solution of this recurrence is $$P(k,n)=\frac{k!}{n!(k-n)!} p^n (1-p)^{k-n}$$
$$Q(k,n)=P(k-1,n-1)p=\frac{(k-1)!}{(n-1)!(k-n)!} p^{n-1} (1-p)^{k-n} p$$
$$\frac{P(k,n)}{Q(k,n)}=\frac{\frac{k!}{n!(k-n)!} p^n (1-p)^{k-n}}{\frac{(k-1)!}{(n-1)!(k-n)!} p^{n-1} (1-p)^{k-n} p}=\frac{k!}{(k-1)!}\frac{(n-1)!}{n!}=\frac{k}{n}$$
$$P(k,n)=\frac{k}{n} Q(k,n)$$
$$P(9,3)=\frac{9}{3} Q(9,3)=3Q(9,3)$$
But situation described in question is impossible, because $Q(9,3)$ cannot be equal 0.2342:
$$Q(9,3)=\frac{8!}{2!\cdot 6!}p^3 (1-p)^6=28 (p(1-p)^2)^3$$
Using AM-GM for $0 \leq p \leq 1$:
$$p(1-p)^2 =\frac{1}{2}\cdot 2p\cdot (1-p)\cdot (1-p) \leq \frac{1}{2}\cdot \left(\frac{2p+
(1-p)+(1-p)}{3}\right)^3=\frac{4}{27}$$
$$Q(9,3) \leq 28 \cdot\left(\frac{4}{27}\right)^3=\frac{1792}{19683}<\frac{1}{10}$$
