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Brums [2.3K]
3 years ago
11

An electronic office product contains 5100 electronic components. Assume that the probability that each component operates witho

ut failure during the useful life of the product is 0.999, and assume that the components fail independently. Approximate the probability that 2 or more of the original 5100 components fail during the useful life of the product. Use normal approximation. Round your answer to 3 decimal places.
Mathematics
1 answer:
AURORKA [14]3 years ago
7 0

Answer:

The probability that 2 or more of the original 5,100 components may fail during the useful life of the product is:

= 0.001

Step-by-step explanation:

Probability of operating without failure = 0.999

The probability of failed component = 0.001 (1 - 0.999)

The number of components of the electronic office product = 5,100

The number of components that may fail, given the above successful operation = 5.1 (0.001 * 5,100).

Therefore, the probability that 2 or more of the original 5,100 components may fail during the useful life of the product = 0.001

Probability of component failure is the likelihood that a component fails during the useful life of the product.  It is expressed as the number of likely failed components divided by the total number of components.  This result can be left in decimal form or expressed as a percentage.

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mamaluj [8]
<h2>Answer:</h2>

(a)

The probability is :  1/2

(b)

The probability is :  1/2

<h2>Step-by-step explanation:</h2>

The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2)   (1,3)    (1,4)    (1,5)

(2,1)   (2,3)   (2,4)   (2,5)

(3,1)   (3,2)   (3,4)   (3,5)

(4,1)   (4,2)   (4,3)   (4,5)

(5,1)   (5,2)   (5,3)   (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

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We know that it could be calculated by using the formula:

P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}

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P(A\bigcap B)=\dfrac{1}{20}

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P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 9

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P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

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Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

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and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

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