A high-tech company receives a shipment of 100 sensitive electronic components from an electronic manufacturing firm.
This shipment will not be acceptable if six or more of the components are faulty. The high-tech company will randomly select a sample of M $\geq$ 1 components from this shipment and if there are no faulty components in this sample of size M, the shipment will be accepted.
What is the smallest value of M that ensures that the probability that the high-tech company will accept an unacceptable shipment is less than 10 percent?
I am stuck on trying to figure out what is p and n since it is not given? If someone can help get me further on this problem that would be great. Thank you. So far I have gotten this far below:
X is a Hypergeometric (100,6,K)
X = # of defective in sample size K
p = rate of defectives
$P(A|U)=∑n=6100−K\frac{\binom{n}{0}\binom{100−n}{K}}{(100K)}P(N=n∣N≥6)$
$P(N=n|N≥6)=\frac{\binom{100}{n}p^n(1−p)^{100−n}}{\sum_{j=6}^{100}\binom{100}{j}(1−p)^{100−j}}$
Your notation is strange and your reasoning is not clear. Why, if the question uses $M$ for the number of components sampled, do you use $K$? What does $p$ have to do with anything? What do you mean by $P(A \mid U)$ when you have not defined $U$ or $A$? Where does $6100$ come from?
If $M$ is the number of components sampled, there can be anywhere from $0$ to $M$ defects. Let the random number of defects be $X$; then $$X \sim \operatorname{Hypergeometric}(n = 100, d, M)$$ where $d$ is the true number of defective items in the shipment. We can easily calculate $$\Pr[X = 0 \mid d, M] = \frac{\binom{d}{0} \binom{100-d}{M-0}}{\binom{100}{M}} = \frac{\binom{100-d}{M}}{\binom{100}{M}}. \tag{1}$$ This is the conditional probability that, given $d$ defects, a sample of size $M$ will be defect-free and will therefore be accepted. If $d \ge 6$, then the shipment is defective. Note that the denominator of $(1)$ does not depend on $d$, and that, for a fixed choice of $M$, the probability of acceptance decreases with increasing $d$--this is intuitively obvious, since the more defects in your shipment, the less likely you are to pull a sample without any defects. Since we want to ensure that the chance of erroneously accepting a defective shipment is less than $0.1$, we may select $d = 6$ as an upper bound: $$\Pr[X = 0 \mid (d \ge 6) \cap M] \le \Pr[X = 0 \mid (d = 6) \cap M] = \frac{\binom{94}{M}}{\binom{100}{M}} = \prod_{i=0}^5 \left( 1 - \frac{M}{100-i} \right). \tag{2}$$ We want to choose the smallest $M$ such that the product in $(2)$ is less than $0.1$. To save computational effort in guessing $M$, we have via the AM-GM identity $$\prod_{i=0}^5 \left( 1 - \frac{M}{100-i} \right) \le \left( \frac{1}{6} \sum_{i=0}^5 1 - \frac{M}{100-i} \right)^6 = \left(1 - M \sum_{i=95}^{100} \frac{1}{i} \right)^6 \approx (1 - 0.0102596 M)^6.$$ Equating this with $0.1$ gives $M \approx 31.0645$, which furnishes our initial guess. Now if we compute the exact probability for values around this guess, we have the table $$\begin{array}{c|c} M & \Pr[X = 0 \mid (d = 6) \cap M] \\ \hline 31 & \frac{681122}{6773025} \approx 0.100564 \\ 32 & \frac{29614}{322525} \approx 0.0918192 \end{array}$$ Therefore, $M = 32$ is the least such $M$.