Question 1172797: In a lot there are 1100 pieces, 19% of them are broken. We choose randomly, successively, without adding them back, 3 pieces from the lot. What's the probability that:
a) 2 out of 3 extracted pieces are broken?
b) Maximum 1 piece is broken?
c) The second piece to be broken?
d) The third piece to be broken?
Answer by VFBundy(438)   (Show Source):
You can put this solution on YOUR website! If there are 1100 pieces in a lot, and 19% of them are broken, this means that exactly 209 pieces are broken (1100 * 0.19 = 209), while the remaining 891 pieces are not broken (1100 - 209 = 891).
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a) 2 out of 3 extracted pieces are broken?
 =  = 0.0875
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b) Maximum 1 piece is broken?
We want to figure out the probability that exactly ONE piece is broken AND the probability that NONE of the pieces are broken.
Probability exactly ONE piece is broken:
 =  = 0.3746
Probability NONE of the pieces are broken:
 =  = 0.5311
Probability maximum one piece is broken:
0.3746 + 0.5311 = 0.9057
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c) The second piece to be broken?
Probability first piece is broken and second piece is broken:
 =  = 0.0360
Probability first piece is NOT broken and second piece is broken:
 =  = 0.1540
Probability second piece is broken:
0.0360 + 0.1540 = 0.1900
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d) The third piece to be broken?
Probability first piece is broken, second piece is broken, and third piece is broken:
 =  = 0.0068
Probability first piece is broken, second piece is NOT broken, and third piece is broken:
 =  = 0.0292
Probability first piece is NOT broken, second piece is broken, and third piece is broken:
 = = 0.0292
Probability first piece is NOT broken, second piece is NOT broken, and third piece is broken:
 =  = 0.1249
Probability third piece is broken:
0.0068 + 0.0292 + 0.0292 + 0.1249 = 0.1901
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