Probability

[mathproblems]

Six microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining at least three defective microprocessors.

((90,6)*(10,3))/(100,6)

Is

this correct? Thank you.

 

[pka]

Six microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining at least three defective microprocessors.

((90,6)*(10,3))/(100,6)

No it is not correct.
At least three means, three, four, five, or six are defective.

 

[mathproblems]

Ok, then to find no defective: (90,6)/(100,6) = 0.6516

at least one is: 1- (90,6)/(100,6)=1-0.6516= 0.3484

and what about at least 3? 0.3484 * 3?

Thank you.

No it is not correct.
At least three means, three, four, five, or six are defective.

 

[soroban]

Hello, mathproblems!

Six microprocessors are randomly selected from a lot of 100 microprocessors of which 10 are defective.
Find the probability of getting at least three defective microprocesors.

There are: .\(\displaystyle {100\choose6}\) possible outcomes.


"At least 3 defective means: .(3 def) or (4 def) or (5 def) or (6 def).

. . 3 def and 3 good: .\(\displaystyle {10\choose3}{90\choose3}\) ways.

. . 4 def and 2 good: .\(\displaystyle {10\choose4}{90\choose2}\) ways.

. . 5 def and 1 good: .\(\displaystyle {10\choose5}{90\choose1}\) ways.

. . 6 def and 0 good: .\(\displaystyle {10\choose6}{90\choose0}\) ways.


Hence, there are: .\(\displaystyle {10\choose3}{90\choose3} + {10\choose4}{90\choose2} + {10\choose5}{90\choose1} + {10\choose6}\) ways

. . . . . . . . . . . . . . to get at least 3 defective items.


\(\displaystyle \displaystyle \text{Therefore: }\(\text{at least 3 d{e}f}) \;=\;\frac{{10\choose3}{90\choose3} + {10\choose4}{90\choose2} + {10\choose5}{90\choose1} + {10\choose6}}{{100\choose6}} \)

 

[mathproblems]

thank you!!