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: 0.14

Explanation:

The expert just didnt round the answer for you guys.

Answer 2

Answer:
`((41)/(50))^(10)`

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 =
`(18)/(100)=(9)/(50)`

⇒ The probability of not finding any bug in first attempt =
`1 - (9)/(50) =(41)/(50)`

Similarly, in second attempt , third attempt, fourth attempt_ _ _ _ _ tenth attempt, the probability of not finding any bug is also equal to
`(41)/(50)`

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

=
`(41)/(50)* (41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)*(41)/(50)`

=
`((41)/(50))^(10)`