SOLUTION: Jane, a quality-control inspector has thirteen assembly lines from which to choose products for testing. Each morning, she randomly selects one of the lines to work on for the day.
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Question 1130394: Jane, a quality-control inspector has thirteen assembly lines from which to choose products for testing. Each morning, she randomly selects one of the lines to work on for the day. Find the probability that no line is chosen more than once for inspection in a span of three days. Answer by ikleyn(52786) (Show Source):
Imagine that all and each line are marked by the first 13 letters of English alphabet:
1 2 3 4 5 6 7 8 9 10 11 12 13
A B C D E F G H I J K L M
Then the space of all possible events is the set of all 3-letter words comprising of these letters.
Repetitions of letters in these words are allowed.
It is easy to calculate the number of all such 3-letter words.
Any of 13 letter can stay in the 1-st position. This gives 13 opportunities.
Any of 13 letter can stay in the 2-nd position. This gives 13 opportunities.
Any of 13 letter can stay in the 3-rd position. This gives 13 opportunities.
In all, there are such words.
Correspondingly, there are elements in the space of events, in all.
Now, the favorable events are those 3-letter words what have no repetitions.
The number of such words is exactly 13*12*11 = 1716.
Therefore, the probability under the question is equal to
= 0.7811 = 78.11% (approximately). ANSWER
