SOLUTION: A box contains 33 parts, of which 3 are defective and 30 are nondefective. 2 parts are selected without replacement. (1) Find the probability P(both are defective). (Give your

Algebra ->  Probability-and-statistics -> SOLUTION: A box contains 33 parts, of which 3 are defective and 30 are nondefective. 2 parts are selected without replacement. (1) Find the probability P(both are defective). (Give your      Log On


   

Question 1000809: A box contains 33 parts, of which 3 are defective and 30 are nondefective. 2 parts are selected without replacement.

(1) Find the probability P(both are defective). (Give your answer correct to two decimal places.)
(2)Find the probability P(exactly one is defective). (Give your answer correct to two decimal places.)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
P%28D%29=3%2F33=1%2F11
P%28D%2CD%29=%281%2F11%29%281%2F11%29=1%2F121
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P%28ND%29=30%2F33=10%2F11
Since you don't know whether the first part is defective or the second, you count both instances.

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.
The last remaining probability is neither part is defective.
P%28ND%2CND%29=%2810%2F11%29%2810%2F11%29=100%2F121
If you sum all those probabilities, you will get 1, which shows you that these are the only three possible occurences.