Probability of 2 Defective Laptops Among 6 Purchased

  • Thread starter r0bHadz
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In summary, the probability of exactly 2 defective laptops among 6 laptop computers, where 5 out of 10 laptops are defective, is the same as the probability of getting two "B"s in a string of 6 letters "G" and "B" with at most 5 "B"s or 5 "G"s. This probability is the same for all strings that contain exactly 2 "B"s, regardless of their specific arrangement.
  • #1
r0bHadz
194
17

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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  • #2
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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  • #3
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?
 
  • #4
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?
 
  • #5
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.
 

Related to Probability of 2 Defective Laptops Among 6 Purchased

What is the probability of getting 2 defective laptops out of 6 purchased?

The probability of getting 2 defective laptops out of 6 purchased can be calculated using the binomial distribution formula. This formula takes into account the number of trials (6), the probability of success (the chance of getting a defective laptop), and the number of successes (2).

How do you calculate the probability of 2 defective laptops out of 6 purchased?

To calculate the probability of 2 defective laptops out of 6 purchased, you can use the binomial distribution formula: P(x) = (nCx)(p^x)(q^(n-x)), where n is the number of trials, p is the probability of success, and q is the probability of failure. In this case, n = 6, p = the probability of getting a defective laptop, and q = 1 - p.

What is the probability of getting 2 or more defective laptops out of 6 purchased?

The probability of getting 2 or more defective laptops out of 6 purchased can be calculated by adding the individual probabilities of getting exactly 2, 3, 4, 5, or 6 defective laptops. You can also use the binomial cumulative distribution function to calculate this probability.

What factors can affect the probability of getting 2 defective laptops out of 6 purchased?

The probability of getting 2 defective laptops out of 6 purchased can be affected by various factors such as the quality control of the manufacturer, the type of laptops purchased, and the condition of the laptops. Additionally, if the laptops were purchased from different sources, the probability may vary depending on the quality of each source.

What can be done to reduce the probability of getting 2 defective laptops out of 6 purchased?

To reduce the probability of getting 2 defective laptops out of 6 purchased, you can research the quality and reliability of the manufacturer and the source from which you are purchasing the laptops. You can also consider purchasing a warranty or insurance for the laptops to protect yourself in case of defects. Additionally, thoroughly inspecting the laptops upon purchase can help identify any potential issues.

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