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Irina18 [472]
3 years ago
15

A computer programmer has an 18% chance of finding a bug in any given program. What is the probability that she does not find a

bug within the first 10 programs she examines?
Mathematics
1 answer:
LiRa [457]3 years ago
5 0

Answer:  (\frac{41}{50})^{10}

Step-by-step explanation:

Since, According to the question,

The chance of finding a bug in a program = 18%

Thus, the probability of finding the bug in first attempt = \frac{18}{100}=\frac{9}{50}

⇒ The probability of not finding any bug in first attempt = 1 - \frac{9}{50} =\frac{41}{50}

Similarly, in second attempt , third attempt, fourth attempt_ _ _ _ _ tenth attempt, the probability of not finding any bug is also equal to \frac{41}{50}

Thus, the probability that she does not find a bug within the first 10 programs she examines

= \frac{41}{50}\times \frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}

= (\frac{41}{50})^{10}

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