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Elanso [62]
3 years ago
15

In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been p

roperly drilled in the boxes and in ten minutes if the holes must be redrilled. Twenty-one gearboxes are in stock, 4 with improperly drilled holes. Five gearboxes must be selected from the 21 that are available for installation in the next five robots. (Round your answers to four decimal places.) (a) Find the probability that all 5 gearboxes will fit properly.
Mathematics
1 answer:
slavikrds [6]3 years ago
7 0

Answer: Our required probability is 0.304.

Step-by-step explanation:

Since we have given that

N = 21

Number of improperly drilled holes k = 4

Number of properly drilled holes = N-k= 21-4=17

Let X be the number of improperly drilled holes in a sample of 5.

We will use "Binomial distribution":

P(X=5)=\dfrac{^4C_0\times ^{17}C_5}{^{21}C_5}=0.304

So, our required probability is 0.304.

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First this problem it should be the group like terms.

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See below

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Permutation is to select an object then arrange it and it cares about the orders while Combination is about only selecting an object without caring the orders.

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where n is a number of total object and r is a number of selected object to arrange. Hence. n cannot be less than r.

Now let's see an example of permutation, suppose we have letter A, B and C. I'd like to know how many ways these words can be arranged:

Since there are 3 letters total and 3 selected letters to arrange then:

\displaystyle{_3 P _3 = \dfrac{3!}{(3-3)!}}\\\\\displaystyle{_3 P _3 = \dfrac{3 \times 2 \times 1}{0!}}\\\\\displaystyle{_3 P _3 = \dfrac{6}{1}}\\\\\displaystyle{_3 P _3 = 6}

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Notice that if you do visually, you'll get the same answer as the calculation of permutation!

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Combination can be expressed mathematically as:

\displaystyle{_n C _r = \dfrac{n!}{(n-r)!r!} = \dfrac{_n P _r}{r!} \ \ \ (n \geq r) }

The difference between permutation and combination is that you only find how many ways you can select object in combination. Therefore, no arrange and doesn't care about order, just ways to select.

Suppose we have same 3 letters: A, B and C. I want to find how many ways I can select these 3 letters:

Since there are 3 letters total and 3 selected letters:

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Hence, there is only one way to select 3 letters. This makes sense because if you have 3 letters then you can only select 3 letters only one way.

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