Question 1162283: A quality-control procedure for testing digital cameras consists of drawing two cameras at random
from a lot of 100 without replacing the first camera before drawing the second. If both are defective,
the entire lot is rejected. If the lot contains 10 defective cameras, assume that each camera in the
lot has an equal chance of being drawn. Answer the following questions.
a) Find the probability of getting a defective camera on the first draw. (2 marks)
b) List TWO possible events from the experiment. (4 marks)
c) Draw a well-labelled tree diagram to show the all the joint probabilities. (10 marks)
d) Are the event “first camera drawn is defective” and the event “second camera drawn is
defective” mutually exclusive? Justify your answer. (3 marks)
e) Are the event “first camera drawn is defective” and the event “second camera drawn is
defective” independent or not? Justify your answer. (5 marks)
f) Use the appropriate multiplication rule to obtain the probability that the entire lot being
rejected. (3 marks)
g) If one of the cameras drawn is defective, then another two cameras will be drawn from the
second lot. Find the probability that the two cameras will be drawn from second lot of 100
cameras. (3 marks)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a. That is 10/100 or 1/10
b. DD DN (others that work are ND and NN)
c.
D(0.1)------D(1/11)
-------N (10/11)
N-(0.9)-----D(10/99)
-------N(89/99)
d. No, because if the first camera is defective the second camera certainly can be.
e. Not independent, because the result of the first changes the probability of the second, not a lot but definitely different.
f. chance 1st is defective is 1/10 and the second is 9/99 or 1/11. That product is the answer and it is 1/110.
g. chance 1st is defective and 2nd is not or 1st is normal and the 2nd defective is 1/10*90/99 (10/11) and 90/100*10/99. That is 1/11+1/11 or 2/11.
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