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 find the complement of the probability that the number of circuit boards inspected is less than or equal to 10.
First, let's find the probability that a randomly selected circuit board is defective. We are given that 2 percent of the circuit boards are defective, so the probability of selecting a defective board is 0.02.
Now, let's find the probability that a randomly selected circuit board is not defective. Since there are only two possibilities (defective or not defective), the probability of selecting a non-defective board is 1 - 0.02 = 0.98.
Next, let's find the probability that the number of circuit boards inspected before a defective board is found is less than or equal to 10. This is the probability of selecting a non-defective board 10 times in a row.
P(inspected before a defective board is found ≤ 10) = (0.98)^10 = 0.8164
Finally, let's find the complement of this probability to get the probability that the number of circuit boards inspected before a defective board is found is greater than 10.
P(inspected before a defective board is found > 10) = 1 - P(inspected before a defective board is found ≤ 10) = 1 - 0.8164 ≈ 0.1836
Therefore, the probability that the number of circuit boards inspected before a defective board is found is greater than 10 is approximately 0.1836 or 18.36%.