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A robotic insertion tool contains 17 primary components. The probability that any component fails during the warranty period is 0.01. Assume that the components fail independently and that the tool fails if any component fails. What is the probability that the tool fails during the warranty period? Round your answer to four decimal places (e.g. 98.7654).

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jacob193

Answer:

0.1571 (rounded to four decimal places as required.)

Step-by-step explanation:

There are two possible ways to find this probability.

Approach one:

find the probability of each of the following cases:

  • One out of the 17 components fails.
  • Two out of the 17 components fail.
  • [tex]\cdots[/tex]
  • All 17 components fail.

The probability that any component fails is equal to the sum of all these 17 probabilities.

Approach two:

Find the probability complement of the event that any of the seventeen components fail. In other words, what is the probability that none of the seventeen components fails?

[tex]P(\text{Any component fails}) = 1 - P(\text{No component fails})[/tex].

This explanation implements the second method.

For each component, the probability that it fails is [tex]0.01[/tex]. The probability that it does not fail is equal to [tex]1 - 0.01 = 0.99[/tex].

The probability that each component fails or does not fail is independent and equal to [tex]0.99[/tex] for all 17 components. As a result, the probability that none of the components fail can be expressed as

[tex]P(\text{No component fails}) = (0.99)^{17}[/tex].

Again,

[tex]\begin{aligned}&P(\text{Any component fails}) \\&= 1 - P(\text{No component fails})\\ &= 1 - (0.99)^{17}\\ &\approx 0.1571 \end{aligned}[/tex]

The final result is rounded to four decimal places as requested.

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