# Sampling with replacement - Probability [closed]

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 $$\frac{(46\text{C}3\times 4\text{C}2)}{50\text{C}5}$$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, José Carlos Santos, Paul Frost, воитель
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Ben Collister Jul 2 at 15:03

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.

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.
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.