A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day for sharpness.
If any dull blade is found, the assembly is replaced with a newly sharpened set of blades.
a. If 10 of the blades in an assembly are dull, what is the probability that the assembly is replaced the first day it is evaluated?
b. If 10 of the blades in an assembly are dull, what is the probability that the assembly is not replaced until the third day of
evaluation? [Hint: Assume that the daily decisions are independent, and use the geometric distribution.]
c. Suppose that on the first day of evaluation, 2 of the blades are dull; on the second day of evaluation, 6 are dull;
and on the third day of evaluation, 10 are dull. What is the probability that the assembly is not replaced until the third day
of evaluation? [Hint: Assume that the daily decisions are independent. However, the probability of replacement changes every day.]
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(a) So, what we have ?
We have an assembly of 48 blades.
10 blades of 48 blades of this assembly are dull.
We check 5 arbitrary blades of 48 (with no replacement).
What is the probability that at least in one trial we will find a dull blade ?
But of course, we should turn the sock inside out and to consider the COMPLEMETNTARY event
and calculate the COMPLEMENTARY probability that no one of 5 tested blade is dull.
The probability that no one of 5 tested blude is dull is
P' = = 0.3856 (rounded).
+--------------------------------------------------------------------+
| You see here the product of 5 specific factors, that represent |
| probabilities for each of 5 consecutive tested blades to be dull. |
+--------------------------------------------------------------------+
We want the complementary value to it. So, our ANSWER is
P = 1 - P' = 1 - = 1 - 0.3856 = 0.6144.
At this point, the solution for part (a) is complete.
(b) Question (b) asks you to find the probability that the given assembly of 48 blades will survive the day 1 and the day 2
(but it does not asks about day 3)
The answer is easy: the probability that this assembly will survive day 1 and will not be replaced after day 1 testing
is 0.3856, as we saw it in part (a).
So, if it survive, this assemble is not replaced and goes "as is" to the test of day 2.
The probability that this assembly will survive day 2 and will not be replaced after day 2 testing
is the same value of 0.3856 (the second day testing is independent on the first day testing).
Therefore, the probability that the given assembly will survive day 1 and day 2 testing is the product
0.3856 * 0.3856 = 0.1487 (rounded). ANSWER
At this point, the solution for part (b) is complete.
Having this method and these templates in front of you, you can easily solve part (c) on your own.
All necessary ideas for it you just have in my solutions to parts (a) and (b).
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As a conclusion, I want to say that the problem's formulation is unsatisfactory.
Indeed, in question (c), it asks "What is the probability that the assembly is not replaced until the third day of evaluation".
The precise (literal) meaning of this question in English is
"What is the probability that the assembly is not replaced to the end of the second day of testing?"
If so, then why in part (c) they talk about some details of activity at the day 3 ?
To me, it tells that the authors either do not know English, or intently try to confuse a reader,
or do not know, at all, what is accurate using words in Math problems.
Or all three of these, simultaneously.