In a Sample of 10 telephones, 4 are defective. If 2 are selected at random and tested, what is the probability that all will be non-defective? This problem is dependent because it matters which one you choose, So i don't think we can do the multiplication thing in this one. 


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*Probability of ( non defective ) = 6/10 


What does the question mean when it says all will be non-defective? is "all" the 2 randomly chosen telephone? How would i do this problem? 2 is chosen randomly and 6 is non defective, I just thought of doing 2/6 cause 2 was the chosen and 6 total are non-defective. But if i wanted to find the number of non defective i would just do 6/10? I feel like I don't understand this question
 A: *

*How many ways are there to choose 2 telephones to test out of the 10 telephones in the sample?

*How many ways are there to choose 2 non-defective telephones out of the 6 non-defective telephones in the sample?
The answers to both of the above can be computed using binomial coefficients. Now, divide the answer to question 2 by the answer to question 1 to get the probability of picking 2 non-defective telephones.
A: Imagine picking the phones one at a time. As you mentioned, the probability the first phone picked is OK is $\frac{6}{10}$. Given that this happened, the probability the second phone is OK is $\frac{5}{9}$, since there are only $5$ good phones left.
Thus our probability is $\frac{6}{10}\cdot \frac{5}{9}$. This product can be simplified.
A: You have 10 phones. This is your "sample". You know (you are told) that exactly 4 are defective. Now, you randomly selected two. The question is asking, "What is the probability that these 2 are not defective?". It is not explicit, but I think you can take it to mean that you do not put the phone back. So each draw is dependent. But you can still multiply, like so: First, you draw one telephone. There is a $6/10$ chance to draw a working phone. On the next draw, there are only 5 working phones left, out of a total of 9, since we just drew a working one. Making this consideration is what is called conditioning, and we use the word "given" and use the symbol "|".Thus, the probability of two working phones is:
$$P(W_1,W_2) = P(W_1)P(W_2|W_1) = \frac{6}{10}\cdot \frac{5}{9} = \frac{30}{90}.$$
This is the product rule, where $W_i = \text{Draw $i$ is working phone}$, for $i = 1,2$.
