Two percent of the circuit boards manufactured by a particular company are defective. If circuit boards are randomly selected for testing, the probability that the number of circuit boards inspected before a defective board is found is greater than 10 is

To find the probability that the number of circuit boards inspected before a defective board is found is greater than 10, we need to calculate the sum of probabilities from getting a non-defective board for the first 10 trials.

Since we know that 2% of the circuit boards are defective, the probability of selecting a non-defective board is 1 - 0.02 = 0.98.

The probability of getting a non-defective board for the first trial is 0.98. For the second trial, the probability is also 0.98.

We can continue this pattern until the tenth trial, finding the probability for each trial, and multiply them together since each trial is independent.

So, the probability of obtaining a non-defective board for the first 10 trials is (0.98)^10.

To find the probability that the number of circuit boards inspected before a defective board is found is greater than 10, we need to find the complement of this probability.

The complement of finding a defective board within 10 trials is finding it after 10 trials or more. So, the probability we need to find is 1 - (0.98)^10.

Now we can calculate this value:

1 - (0.98)^10 ≈ 1 - 0.8165 ≈ 0.1835

Therefore, the probability that the number of circuit boards inspected before a defective board is found is greater than 10 is approximately 0.1835.