Question

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?

Answer 1

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

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

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