document.write( "Question 1162283: A quality-control procedure for testing digital cameras consists of drawing two cameras at random
\n" ); document.write( "from a lot of 100 without replacing the first camera before drawing the second. If both are defective,
\n" ); document.write( "the entire lot is rejected. If the lot contains 10 defective cameras, assume that each camera in the
\n" ); document.write( "lot has an equal chance of being drawn. Answer the following questions.
\n" ); document.write( "a) Find the probability of getting a defective camera on the first draw. (2 marks)
\n" ); document.write( "b) List TWO possible events from the experiment. (4 marks)
\n" ); document.write( "c) Draw a well-labelled tree diagram to show the all the joint probabilities. (10 marks)
\n" ); document.write( "d) Are the event “first camera drawn is defective” and the event “second camera drawn is
\n" ); document.write( "defective” mutually exclusive? Justify your answer. (3 marks)
\n" ); document.write( "e) Are the event “first camera drawn is defective” and the event “second camera drawn is
\n" ); document.write( "defective” independent or not? Justify your answer. (5 marks)
\n" ); document.write( "f) Use the appropriate multiplication rule to obtain the probability that the entire lot being
\n" ); document.write( "rejected. (3 marks)
\n" ); document.write( "g) If one of the cameras drawn is defective, then another two cameras will be drawn from the
\n" ); document.write( "second lot. Find the probability that the two cameras will be drawn from second lot of 100
\n" ); document.write( "cameras. (3 marks)
\n" ); document.write( "

Algebra.Com's Answer #786071 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
a. That is 10/100 or 1/10
\n" ); document.write( "b. DD DN (others that work are ND and NN)
\n" ); document.write( "c.
\n" ); document.write( "D(0.1)------D(1/11)
\n" ); document.write( "-------N (10/11)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "N-(0.9)-----D(10/99)
\n" ); document.write( "-------N(89/99)\r
\n" ); document.write( "\n" ); document.write( "d. No, because if the first camera is defective the second camera certainly can be.
\n" ); document.write( "e. Not independent, because the result of the first changes the probability of the second, not a lot but definitely different.
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( "
\n" );