Question 876990: a delivery truck company has 20 trucks, three of which have faulty brakes. if three of the 20 trucks are randomly selected, what is the approximate probability that the first two trucks have good brakes, and the third truck has faulty brakes?
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Use the formula for compound events without replacement.
The probability of first selecting a good tire is 17/20. Then the probability of the second selection being good is 16/19, because we assume the fisrt event occurred corectly, leaving only 16 good tires in a total of only 19 tires because the first selection was not replaced. The probability of the third selection being faulty is 3/18, because the first two selections were not faulty, so all three faulty tires are in the total sample of only 18 tires because the first two selection were not replaced. This gives us
(1) P(GGF) = (17/20)*(16/19)*(3/18) or
(2) P(GGF) = 816/6840 or
(3) P(GGF) = 0.119
Note that the denominator of (2) is 20P3, the permutations of 20 items taken 3 at a time or the total number of our sample size.
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