Probability

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
An assembly line produces 60 motors, including 10 that are defective. The quality control
department selects a random sample of 4 of the motors.


(a) What is the probability that exactly 3 of the motors in the sample are defective?
(b) What is the probability that at least one of the motors in the sample is defective?

Im pretty sure the answers are this:

a) 0.012304284, which is 1.2 % ( To get this I choose 3 from the defective, and one good item, which is (10/3) times (50/1) divided by (60/4) (since that was the total batch of motors, and you choose a sample of 4)

b) 0.000020507, because you do (10/1) times (50/0) to get 10, then you take 10 and divide it into (60/4).

Not 100% on if these right or not, but I think thats how you do it!
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Helloo, quaidy4!

Your first answer is correct, and so is your reasoning . . . Good work!

But your second answer is wrong.


(b) What is the probability that at least one of the motors in the sample is defective?

There are 60 motors: 10 Defective and 50 Good.

"At least one Defective" means: .1 defective or 2 defectives or 3 defectives or 4 defectives.
. . You must compute the four probabilities and add them.


It is easier to find the opposite probability: no defectives.
. . This means all 4 motors are Good.

\(\displaystyle \text{Choose 4 of the 50 Good motors: }_{50}C_4\text{ ways.}\)

\(\displaystyle \text{Divide by the number of possible samples: }_{60}C_4\)

\(\displaystyle \text{Hence: }\:p(\text{no Defective}) \;=\;\frac{_{50}C_4}{_{60}C_4} \;=\;\frac{46,\!060}{97,\!527}\)


\(\displaystyle \text{Therefore: }\:p(\text{at least one Defective}) \;\;=\;\;1 - \frac{46,\!060}{97,\!527} \;\;=\;\;\frac{51,\!467}{97,\!527} \;\;\approx\;\;52.8\%\)

 

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
oh okay! thank you so much! :)
 
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