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Technical control checks a series of 100 devices. Device has type A defect with the probability of $0.01$ and type B defect with probability of $0.02$ these two defects are independent (type A wont increase or decrease a chances of appearing type B defect and vice versa). They conclude that device if defected if it has at least on of these two defects. Find the probability that there are two to five defected devices.

Poisson distribution:

$$P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}$$

Lets say that $X$ is the number of defected devices, we can see that Poisson distribution will do just fine here since $np_A=1$ and $np_B=2$, but thats not the problem.

We can say that there can be two devices with defect A and that can be easily calculated, however, it might happen that there are two devices defected with type B defect but it can also happen that BUT, it can also happen that there are one with A and one with B, question is, should this case need to be counted twice, since for example, 55. device might have defect A and 88. can have defect B and thats one case, since it can happen other way around should that be calculated twice or not?

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Hint:

For a device let $A$ denote the event that it has type A defect and $B$ the event that it has type B defect.

Further let $D$ denote the event that it has at least one defect.

Then: $$P(D)=P(A\text{ or } B)=P(A)+P(B)-P(A\text{ and }B)=P(A)+P(B)-P(A)P(B)$$

The second equality is based on independence.

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  • $\begingroup$ So actually i have to find the probability that two devices have defect A, then do the same for B, and then use inclusion-exclusion formula, to get the probability for X=2. Then after that, i should repeat the same procedure for X=3,4,5.Is that right? $\endgroup$ – cdummie May 7 '18 at 10:26
  • $\begingroup$ Wait. Not for two devices, but just for one you should find the probability whether it is defect or not. From there you can work with $p_D$ in stead of $p_A,p_B$. $\endgroup$ – drhab May 7 '18 at 10:30
  • $\begingroup$ But it says "Find the probability that there are two to five defected devices." why should i care for probability of one device being defected? $\endgroup$ – cdummie May 7 '18 at 10:39
  • $\begingroup$ The probability that one device is defect gives you the correct parameter to work with. $\sum_{k=2}^5\binom{100}{k}p_D^k(1-p_D)^{100-k}$ or (with Poisson) $e^{-np_D}\sum_{k=2}^5\frac{(np_D)^k}{k!}$ is the probability that $2,3,4$ or $5$ devices are defect. $\endgroup$ – drhab May 7 '18 at 11:57
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Poisson distribution will not do just fine, since there are only a 100 devices, and yet $P(X = 101) > 0$.

Think of the following questions as a hint:

a) What is the probability that a device has a defect?

b) Given that each device is independent, and there are a 100 devices, what is the probability that exactly 2 of them have a defect?

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    $\begingroup$ I think Poisson is "good enough" here (relatively large $n$ and small $p$). $\endgroup$ – drhab May 7 '18 at 8:22
  • $\begingroup$ I agree, but it should probably be clarified by either where the OP got the question from (something like "use Poisson approximation to compute") or by OP themselves to avoid confusion. If not specified, I would definitely leave it as a sum of bunch of terms :D $\endgroup$ – E-A May 7 '18 at 8:24
  • $\begingroup$ I probably should have noted that, as far as i know, Poisson distribution is fine as long as the product $np$ (n number of elements, p probability for one of them) is less than 10, now, that might not be precise enough if you are advanced in probability, this in my case is, however, an introductory course. $\endgroup$ – cdummie May 7 '18 at 10:22
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The probability that no defect is detected is $0.99\cdot0.98$. Now let $X$ denote the number of detected defective devices. Then $X$ follows a binomial distribution with $n=100$ and $p=1-0.99\cdot 0.98$. Now compute $P(2\leq X\leq5)$.

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