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There are 4 failures among $50$ products. You randomly select $5$ samples among these products.

(1) If you sampled $5$ without replacement, what is the probability that exactly $2$ of them are failure?

(2) If you sampled $5$ with replacement, what’s the probability that exactly $2$ of them are failure? The order does not matter.

I need help with the second one. How do i apply this using the method for combinations? I get that the number of total samples stays 50 every time but cannot seem to understand what that does. For the first one, I did $\frac{(46\text{C}3\times 4\text{C}2)}{50\text{C}5}$

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closed as off-topic by Shailesh, mrtaurho, José Carlos Santos, Paul Frost, воитель Jul 2 at 21:54

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    $\begingroup$ Personally, I would model the second part using a Binomial random variable. Also, this sounds similar to the problem of getting two heads out of 3 coin tosses for example, so consider what techniques you would apply to that problem. $\endgroup$ – Ben Collister Jul 2 at 15:03
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Using Ben Collister's suggestion, if the probability of a single unit failing is $p$, then if you choose $n$ units randomly (with replacement), the probability of exactly $k$ failing is given by:

$$({_n}C_k)p^k(1-p)^{n-k}$$

Plugging in $n=5, k=2, p=\dfrac{4}{50}$ will give you your answer.

Answer, about 4.98%.

The idea: if you want to represent a sample, you have each chosen item as either a success or a failure. Each item that is a success occurs with probability $\dfrac{46}{50}$ while each failure occurs with probability $\dfrac{4}{50}$. You want three successes, so you have a factor of $\left(\dfrac{46}{50}\right)^3$. You have two failures, so you have a factor of $\left(\dfrac{4}{50}\right)^2$. Then, you multiply by every possible order of getting two failures in a group of 5. For example: SSSFF SSFSF ...

would be one way to represent the order of successes and failures. You need a string of five characters, two F's and three S's. You can choose the position of the two F's and the rest of the characters are S's. There are ${_5}C_2$ ways of choosing the position of the F's, so that is how many orders there are.

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If you sample with replacement, you know that with each draw, you are picking from the exact same set of items (i.e. $4$ defective ones from $50$ total). From here, you can determine the probability of drawing exactly $2$ defective items in $5$ draws in a particular order, and then figure out in how many different orders you can do this.

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