Probability of 2 Defective Laptops Among 6 Purchased

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The discussion revolves around calculating the probability of selecting exactly 2 defective laptops from a total of 10 laptops, where 5 are defective and 5 are good. Participants explore the relationship between the probabilities of choosing defective laptops and the overall distribution of good and defective laptops. They emphasize the importance of understanding the sample space and the combinations of good and defective laptops. The realization that the probabilities of different arrangements of laptops are the same is highlighted as a key insight for solving the problem. This understanding aids in approaching the probability calculation more effectively.
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Homework Statement


Among 10 laptop computers, five are good and five have defects. Unaware of this, a
customer buys 6 laptops

What is the probability of exactly 2 defective laptops among them?

Homework Equations

The Attempt at a Solution


I'm having a hard time looking at this problem from the definitions in my book.

It seems to me that P(2/6 chosen laptops are defective) is not independent of P(5/10 laptops are defective),
but P(5/10 laptops are defective) is independent of P(2/6) chosen laptops, which would be impossible. This implies that I am looking at this problem from the wrong perspective

also it seems to me, that 1/6 of the chosen laptops, should be defective with probability .5
 
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How many ways of picking 2 defective and 4 working laptops from the 10? How many ways to pick 6 laptops regardless of defective status.
 
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r0bHadz said:

Homework Statement


Among 10 laptop computers, five are good and five have defects. Unaware of this, a
customer buys 6 laptops

What is the probability of exactly 2 defective laptops among them?

Homework Equations

The Attempt at a Solution


I'm having a hard time looking at this problem from the definitions in my book.

It seems to me that P(2/6 chosen laptops are defective) is not independent of P(5/10 laptops are defective),
but P(5/10 laptops are defective) is independent of P(2/6) chosen laptops, which would be impossible. This implies that I am looking at this problem from the wrong perspective

also it seems to me, that 1/6 of the chosen laptops, should be defective with probability .5

The sample-space consists of all strings of 6 letters "G" and "B" with at most 5 "B"s or 5 "G"s. So, a string such as GGBGGG means that his first two items are good, the third is bad, and the remaining three are good, while BBBBBG means the first 5 are bad and the 6th is good, etc. Remember, these are chosen from a set of 5 "G"s and 5 "B"s.

The event E you are interested in is the collection of such strings containing exactly 2 "B"s.

Look at some of these strings: (i) what is P(BBGGGG)? (ii) what is P(GGGGBB)? (iii) what is P(GGBGGB)?

What do you notice about these three probabilities?
 
Ray Vickson said:
The sample-space consists of all strings of 6 letters "G" and "B" with at most 5 "B"s or 5 "G"s. So, a string such as GGBGGG means that his first two items are good, the third is bad, and the remaining three are good, while BBBBBG means the first 5 are bad and the 6th is good, etc. Remember, these are chosen from a set of 5 "G"s and 5 "B"s.

The event E you are interested in is the collection of such strings containing exactly 2 "B"s.

Look at some of these strings: (i) what is P(BBGGGG)? (ii) what is P(GGGGBB)? (iii) what is P(GGBGGB)?

What do you notice about these three probabilities?
I am not the OP but arent these propabilities all the same?
 
Leo Consoli said:
I am not the OP but arent these propabilities all the same?
Yes, indeed, and realization of that fact goes a long way towards solving the problem. I was attempting to help the OP find the solution himself, instead of handing him a completer answer.
 
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