1. ## Failure devices prob.

There are 30 devices, it is known that 20 of them are failure and 10 are ok. There are chosen 4 of them. What is the probability that the chosen devices are not all of them failure/ok?
res. p = 0.611

OK so I tried finding the prob. that 4 of them are failure. I think it is $p^4=(\frac{2}{3})^4=0.197531.$ So the required prob. must be $p=1-p^4=0.802469$. Is it wrong? hehe obviously the res. is different

2. I like your p value. You should consider the binomial theorem here, it is:

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case $n = 30$, $p = \frac{2}{3}$ and $k = 4$

now substituting in these values we get

$P(X=4) = \binom{30}{4} \left(\frac{2}{3}\right)^4\left(1- \frac{2}{3}\right)^{30-4}$

Can you finish it from here?

This will be the probabilty that 4 devices fail, I think that is what the question is asking?

3. Originally Posted by pickslides
I like your p value. You should consider the binomial theorem here, it is:
$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
pickslides,
The binomial theorem is not applicable here. These are not independent trials.
The answer for not all failures is $1-\frac{\binom{20}{4}}{\binom{30}{4}}$.

4. Originally Posted by Plato
pickslides,
The binomial theorem is not applicable here. These are not independent trials.
The answer for not all failures is $1-\frac{\binom{20}{4}}{\binom{30}{4}}$.
I tried using binomial but with n=4, and that leaded me to p^4
Ok Plato some explanations would be appreciated, and why it doesn't fit with the result in my book?

5. Originally Posted by pickslides
I like your p value. You should consider the binomial theorem here, it is:

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case $n = 30$, $p = \frac{2}{3}$ and $k = 4$

now substituting in these values we get

$P(X=4) = \binom{30}{4} \left(\frac{2}{3}\right)^4\left(1- \frac{2}{3}\right)^{30-4}$

Can you finish it from here?

This will be the probabilty that 4 devices fail, I think that is what the question is asking?
No the question is to find the probability that not all of them(4) fail/are ok.

6. Originally Posted by javax
Ok Plato some explanations would be appreciated, and why it doesn't fit with the result in my book?
It is not at all clear what you are asking.
If the question is, "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
The probabillity that all are defective is $\frac{\binom{20}{4}}{\binom{30}{4}}$.

The probabillity that all are non-defective is $\frac{\binom{10}{4}}{\binom{30}{4}}$.

Now note that those two are disjoint events.

The probabillity that all are defective OR all are non-defective is $\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

The probabillity that not all are defective and not all are non-defective is
$1-\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

7. Originally Posted by Plato
It is not at all clear what you are asking.
If the question is, "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
The probabillity that all are defective is $\frac{\binom{20}{4}}{\binom{30}{4}}$.

The probabillity that all are non-defective is $\frac{\binom{10}{4}}{\binom{30}{4}}$.

Now note that those two are disjoint events.

The probabillity that all are defective OR all are non-defective is $\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

The probabillity that not all are defective and not all are non-defective is
$1-\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.
Mate the question is exactly how you said. Your answer still not fitting the given result but I trust you. Thanks

8. Plato one more question.
As I mentioned I tried to find it using $1-p_1^4*p2^4,$ where $p_1=\frac{2}{3}$ and $p_2=\frac{1}{3}$ which is quite close to your result. Is it ok finding it like this?

9. Originally Posted by javax
Mate the question is exactly how you said. Your answer still not fitting the given result but I trust you.
It may be a matter of translation. We may not be doing the same question.
On the other hand, your textbook may be simply be wrong.

10. "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
you fully understood my question. It's ok I believe the given result is wrong (Y)

11. Originally Posted by javax
Plato one more question.
As I mentioned I tried to find it using $1-p_1^4*p2^4,$ where $p_1=\frac{2}{3}$ and $p_2=\frac{1}{3}$ which is quite close to your result. Is it ok finding it like this?
Unless the trials, the events, are independent the binomial formula is not applicable.
Selecting four objects from thirty is in no way independent.

But say we have batch twenty black balls and ten white balls.
We pick a random ball from that collection. Record its color and return it to the batch.
We do that four times. Those outcomes are independent.