Probability of 2 Defective Laptops Among 6 Purchased

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SUMMARY

The probability of selecting exactly 2 defective laptops from a total of 6 purchased out of 10 laptops (5 good and 5 defective) can be calculated using combinatorial methods. The relevant formula involves determining the number of ways to choose 2 defective laptops from 5 and 4 good laptops from 5, divided by the total ways to choose 6 laptops from 10. This results in a probability of 0.5 for selecting 2 defective laptops, as the sample space consists of all combinations of 6 laptops formed from the available good and defective units.

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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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