Among 10 laptop computers 5 are good and 5 are defective. Unaware of this, a customer buys 6 laptops. Given that at least 2 purchased laptops are defective, what is the probability that exactly two are defective?
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Total laptops = 10, good = 5, defective = 5. he bought 6 laptops.
The number of ways of selecting 6 laptops out of 10:
10C2
= 10!/6! * 4!
= 10 * 9 * 8 * 7/4 * 3 * 2 * 1
= 210.
Given 2 purchased laptops are defective:
5C2 * 5C4 = 50
5C2 = 5!/2! * 3!
= 5 * 4 * 3 * 2 * 1/3 * 2 * 1
= 10.
5C4 = 5!/4!*1!
= 5 * 4 * 3 * 2 * 1/ 4 * 3 * 2 * 1 * 1
= 5.
Probability that exactly 2 are defective = 50/210
= 5/21.
Hope this helps!
The number of ways of selecting 6 laptops out of 10:
10C2
= 10!/6! * 4!
= 10 * 9 * 8 * 7/4 * 3 * 2 * 1
= 210.
Given 2 purchased laptops are defective:
5C2 * 5C4 = 50
5C2 = 5!/2! * 3!
= 5 * 4 * 3 * 2 * 1/3 * 2 * 1
= 10.
5C4 = 5!/4!*1!
= 5 * 4 * 3 * 2 * 1/ 4 * 3 * 2 * 1 * 1
= 5.
Probability that exactly 2 are defective = 50/210
= 5/21.
Hope this helps!
spTheTutor:
thank you but this is the solution for finding the probability of exactly two defective laptops. The question wants us to find the probability of exactly two defective laptops 'given that at least two are defective'
Answered by
1
1/5is probability ...........
The answer is 10/41
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